\displaystyle \textbf{Question 1: } \text{In the given figure, }O\text{ is the centre of the circle.}
\displaystyle \angle OAB=30^\circ\text{ and }\angle OCB=40^\circ.\text{ Find }\angle AOC. c11\displaystyle \text{Answer:}
\displaystyle \text{The given solution is incorrect as written due to an erroneous intermediate step.}
\displaystyle \text{However, the final answer }140^\circ\text{ is correct.}
\displaystyle \text{Since }OA=OB,\ \triangle AOB\text{ is isosceles.}
\displaystyle \therefore \angle OBA=\angle OAB=30^\circ
\displaystyle \angle AOB=180^\circ-(30^\circ+30^\circ)=120^\circ
\displaystyle \text{Also, }OB=OC,\ \triangle BOC\text{ is isosceles.}
\displaystyle \therefore \angle OBC=\angle OCB=40^\circ
\displaystyle \angle BOC=180^\circ-(40^\circ+40^\circ)=100^\circ
\displaystyle \text{Angles around point }O\text{ add up to }360^\circ.c11x
\displaystyle \angle AOC+\angle AOB+\angle BOC=360^\circ
\displaystyle \angle AOC+120^\circ+100^\circ=360^\circ
\displaystyle \therefore \angle AOC=140^\circ
\\

\displaystyle \textbf{Question 2: } \text{In the given figure, }\angle BAD=65^\circ,\ \angle ABD=70^\circ
\displaystyle \text{and }\angle BDC=45^\circ.
\displaystyle \text{(i) Prove that }AC\text{ is a diameter of the circle.}
\displaystyle \text{(ii) Find }\angle ACB.\hfill \text{[ICSE 2013]} c12\displaystyle \text{Answer:}
\displaystyle \text{The problem statement is correct and complete.}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{Given, }\angle BAD=65^\circ,\ \angle ABD=70^\circ\text{ and }\angle BDC=45^\circ.
\displaystyle \text{(i) In }\triangle ABD,
\displaystyle \angle BAD+\angle ABD+\angle ADB=180^\circ
\displaystyle 65^\circ+70^\circ+\angle ADB=180^\circ
\displaystyle \angle ADB=180^\circ-135^\circ=45^\circ
\displaystyle \angle ADC=\angle ADB+\angle BDC
\displaystyle \angle ADC=45^\circ+45^\circ=90^\circ
\displaystyle \therefore AC\text{ is a diameter of the circle.}
\displaystyle \text{(Angle in a semicircle is a right angle.)}
\displaystyle \text{(ii) }\angle ACB=\angle ADB
\displaystyle \text{(Angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore \angle ACB=45^\circ
\\

\displaystyle \textbf{Question 3: } \text{In the given figure, }O\text{ is the centre of the circle,}
\displaystyle C,O,B\text{ are collinear and }\angle AOB=70^\circ.
\displaystyle \text{Calculate: (i) }\angle OCA\qquad\text{(ii) }\angle OAC. c13

\displaystyle \text{Answer:}
\displaystyle \text{The original problem is solvable from the figure but is slightly incomplete in words.}
\displaystyle \text{It should explicitly state that }C,O,B\text{ are collinear.}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{Observation: The calculation in }\triangle AOB\text{ is unnecessary.}
\displaystyle \text{Since }C,O,B\text{ are collinear, }\angle AOC+\angle AOB=180^\circ.
\displaystyle \angle AOC+70^\circ=180^\circ
\displaystyle \therefore \angle AOC=110^\circ
\displaystyle OA=OC\qquad\text{(Radii of the same circle)}
\displaystyle \therefore \triangle AOC\text{ is isosceles.}
\displaystyle \angle OCA=\angle OAC
\displaystyle \angle OCA+\angle OAC+\angle AOC=180^\circ
\displaystyle 2\angle OCA+110^\circ=180^\circ
\displaystyle 2\angle OCA=70^\circ
\displaystyle \angle OCA=35^\circ
\displaystyle \text{(i) }\therefore \angle OCA=35^\circ
\displaystyle \text{(ii) }\therefore \angle OAC=35^\circ
\\

\displaystyle \textbf{Question 4: } \text{In each figure, }O\text{ is the centre of the circle.}
\displaystyle \text{Find the values of }a,\ b\text{ and }c.

(i) c141 (ii) c142

\displaystyle \text{Answer:}
\displaystyle \text{The problem statement is correct and complete.}
\displaystyle \text{The given solution is correct, but its reasoning should be stated more clearly.}
\displaystyle \text{(i) The angle at the centre is twice the angle at the circumference}
\displaystyle \text{subtended by the same minor arc.}
\displaystyle 2b=130^\circ
\displaystyle \therefore b=65^\circ
\displaystyle \text{The reflex angle subtended by the major arc is }360^\circ-130^\circ=230^\circ.
\displaystyle 2a=230^\circ
\displaystyle \therefore a=115^\circ
\displaystyle \text{Alternatively, }a+b=180^\circ\text{ since opposite angles of a cyclic}
\displaystyle \text{quadrilateral are supplementary.}
\displaystyle \therefore a=180^\circ-65^\circ=115^\circ
\displaystyle \text{(ii) The reflex angle at }O\text{ is }360^\circ-112^\circ=248^\circ.
\displaystyle 2c=248^\circ
\displaystyle \therefore c=124^\circ
\displaystyle \text{Hence, }a=115^\circ,\quad b=65^\circ\quad\text{and}\quad c=124^\circ.
\\

\displaystyle \textbf{Question 5: } \text{In each figure, }O\text{ is the centre of the circle.}
\displaystyle \text{Find the values of }a,\ b,\ c\text{ and }d.\hfill\text{[ICSE 2007]}

(i)c151 (ii)c152
(iii)c153 (iv)c154

\displaystyle \text{Answer:}
\displaystyle \text{The problem statement is correct and complete when read with the figures.}
\displaystyle \text{The given solution is correct and complete, but some reasons need clarification.}
\displaystyle \text{(i) Since }BD\text{ is a diameter, }\angle DAB=90^\circ.
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle ADB=180^\circ-90^\circ-35^\circ=55^\circ
\displaystyle \angle ACB=\angle ADB
\displaystyle \text{(Angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore a=55^\circ
\displaystyle \text{(ii) Let chords }AC\text{ and }DB\text{ intersect at }E.
\displaystyle \angle BEC=180^\circ-120^\circ=60^\circ
\displaystyle \text{In }\triangle BEC,
\displaystyle \angle BCE=180^\circ-60^\circ-25^\circ=95^\circ
\displaystyle \therefore \angle ACB=95^\circ
\displaystyle \angle ADB=\angle ACB
\displaystyle \text{(Angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore b=95^\circ
\displaystyle \text{(iii) }\angle AOB=2\angle ACB
\displaystyle \angle AOB=2\times50^\circ=100^\circ
\displaystyle OA=OB\qquad\text{(Radii of the same circle)}
\displaystyle \therefore \angle OAB=\angle OBA=c
\displaystyle 2c+100^\circ=180^\circ
\displaystyle 2c=80^\circ
\displaystyle \therefore c=40^\circ
\displaystyle \text{(iv) Since }AB\text{ is a diameter, }\angle APB=90^\circ.
\displaystyle \text{In }\triangle APB,
\displaystyle \angle PAB=180^\circ-90^\circ-45^\circ=45^\circ
\displaystyle \angle PCB=\angle PAB
\displaystyle \text{(Angles in the same segment subtended by chord }PB\text{)}
\displaystyle \therefore d=45^\circ
\displaystyle \text{Hence, }a=55^\circ,\quad b=95^\circ,\quad c=40^\circ,\quad d=45^\circ.
\\

\displaystyle \textbf{Question 6: } \text{In the figure, }AB\text{ is the common chord of the two circles.}
\displaystyle \text{If }AC\text{ and }AD\text{ are diameters, prove that }D,\ B\text{ and }C\text{ are collinear.}
\displaystyle O_1\text{ and }O_2\text{ are the centres of the two circles.} c16\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete.}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{Since }AD\text{ is a diameter, }\angle DBA=90^\circ.
\displaystyle \text{(Angle in a semicircle)}
\displaystyle \text{Also, since }AC\text{ is a diameter, }\angle ABC=90^\circ.
\displaystyle \text{(Angle in a semicircle)}
\displaystyle \angle DBC=\angle DBA+\angle ABC
\displaystyle \angle DBC=90^\circ+90^\circ=180^\circ
\displaystyle \therefore D,\ B\text{ and }C\text{ are collinear.}
\\

\displaystyle \textbf{Question 7: } \text{In the given figure, }AB\parallel DC\text{ and }\angle DAB=105^\circ.
\displaystyle \text{Find: (i) }\angle BCD\qquad\text{(ii) }\angle ADC\qquad\text{(iii) }\angle ABC. c17\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem is correct and complete when read with the figure.}
\displaystyle \text{The given solution is incomplete because part (ii) has not been solved.}
\displaystyle \text{It also uses }\angle ADC=75^\circ\text{ in part (iii) without first proving it.}
\displaystyle \text{The answers obtained for parts (i) and (iii) are correct.}
\displaystyle \text{(i) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle BCD+\angle BAD=180^\circ
\displaystyle \angle BCD+105^\circ=180^\circ
\displaystyle \therefore \angle BCD=75^\circ
\displaystyle \text{(ii) Since }AB\parallel DC\text{ and }AD\text{ is a transversal,}
\displaystyle \angle DAB+\angle ADC=180^\circ
\displaystyle 105^\circ+\angle ADC=180^\circ
\displaystyle \therefore \angle ADC=75^\circ
\displaystyle \text{(iii) Opposite angles of a cyclic quadrilateral are supplementary.}
\displaystyle \angle ADC+\angle ABC=180^\circ
\displaystyle 75^\circ+\angle ABC=180^\circ
\displaystyle \therefore \angle ABC=105^\circ
\displaystyle \text{Hence, (i) }\angle BCD=75^\circ,\quad\text{(ii) }\angle ADC=75^\circ
\displaystyle \text{and (iii) }\angle ABC=105^\circ.
\\

\displaystyle \textbf{Question 8: } \text{In the given figure, }O\text{ is the centre of the circle.}
\displaystyle \text{If }\angle AOB=140^\circ\text{ and }\angle OAC=50^\circ,\text{ find:}
\displaystyle \text{(i) }\angle ACB\qquad\text{(ii) }\angle OBC\qquad\text{(iii) }\angle OAB\qquad\text{(iv) }\angle CBA c18\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete.}
\displaystyle \text{The given solution is partially correct but incomplete in presentation.}
\displaystyle \text{It does not answer the four parts separately and includes an unnecessary step.}
\displaystyle \text{The reflex }\angle AOB=360^\circ-140^\circ=220^\circ.
\displaystyle \text{(i) Angle at the centre is twice the angle at the circumference}
\displaystyle \text{subtended by the same arc.}
\displaystyle \angle ACB=\frac{1}{2}\times220^\circ=110^\circ
\displaystyle \text{(iii) In }\triangle AOB,\ OA=OB\text{ (radii).}
\displaystyle \angle OAB=\angle OBA=\frac{180^\circ-140^\circ}{2}=20^\circ
\displaystyle \angle CAB=\angle OAC-\angle OAB=50^\circ-20^\circ=30^\circc18x
\displaystyle \text{(iv) In }\triangle ACB,
\displaystyle \angle CBA=180^\circ-110^\circ-30^\circ=40^\circ
\displaystyle \text{(ii) }\angle OBC=\angle OBA+\angle ABC
\displaystyle \angle OBC=20^\circ+40^\circ=60^\circ
\displaystyle \therefore \text{(i) }\angle ACB=110^\circ,\ \text{(ii) }\angle OBC=60^\circ,
\displaystyle \text{(iii) }\angle OAB=20^\circ,\ \text{(iv) }\angle CBA=40^\circ.
\\

\displaystyle \textbf{Question 9: } \text{Calculate: (i) }\angle CDB\qquad\text{(ii) }\angle ABC
\displaystyle \text{(iii) }\angle ACB. c19\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete when read with the figure.}
\displaystyle \text{The given solution is partially incorrect.}
\displaystyle \text{The values of }\angle CDB\text{ and }\angle ABC\text{ have been interchanged.}
\displaystyle \text{(i) }\angle CDB=\angle CAB
\displaystyle \text{(Angles in the same segment subtended by chord }CB\text{)}
\displaystyle \angle CAB=\angle BDC=43^\circ
\displaystyle \therefore \angle CDB=43^\circ
\displaystyle \text{(ii) From the figure, }\angle ABC=49^\circ
\displaystyle \text{(iii) In }\triangle ABC,
\displaystyle \angle CAB+\angle ABC+\angle ACB=180^\circ
\displaystyle 43^\circ+49^\circ+\angle ACB=180^\circ
\displaystyle \angle ACB=180^\circ-92^\circ=88^\circ
\displaystyle \therefore \text{(i) }\angle CDB=43^\circ,\quad\text{(ii) }\angle ABC=49^\circ
\displaystyle \text{and (iii) }\angle ACB=88^\circ.
\\

\displaystyle \textbf{Question 10: } \text{In the given figure, }ABCD\text{ is a cyclic quadrilateral}
\displaystyle \text{in which }\angle BAD=75^\circ,\ \angle ABD=58^\circ\text{ and }\angle ADC=77^\circ.
\displaystyle \text{Find: (i) }\angle BDC\qquad\text{(ii) }\angle BCD\qquad\text{(iii) }\angle BCA. c110\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete.}
\displaystyle \text{The given solution is partially correct but incomplete in presentation.}
\displaystyle \text{The calculation for part (ii) is misplaced, and part (iii) is not stated clearly.}
\displaystyle \text{(i) In }\triangle ABD,
\displaystyle \angle ADB=180^\circ-\angle BAD-\angle ABD
\displaystyle \angle ADB=180^\circ-75^\circ-58^\circ=47^\circ
\displaystyle \angle ADC=\angle ADB+\angle BDC
\displaystyle 77^\circ=47^\circ+\angle BDC
\displaystyle \therefore \angle BDC=30^\circ
\displaystyle \text{(ii) Opposite angles of a cyclic quadrilateral are supplementary.}
\displaystyle \angle BAD+\angle BCD=180^\circc110x
\displaystyle 75^\circ+\angle BCD=180^\circ
\displaystyle \therefore \angle BCD=105^\circ
\displaystyle \text{(iii) }\angle BCA=\angle BDA
\displaystyle \text{(Angles in the same segment subtended by chord }BA\text{)}
\displaystyle \therefore \angle BCA=47^\circ
\displaystyle \text{Hence, (i) }\angle BDC=30^\circ,\quad\text{(ii) }\angle BCD=105^\circ
\displaystyle \text{and (iii) }\angle BCA=47^\circ.
\\

\displaystyle \textbf{Question 11: } \text{In the given figure, }O\text{ is the centre and}
\displaystyle \triangle ABC\text{ is equilateral. Find: (i) }\angle ADB\qquad\text{(ii) }\angle AEB. c21\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete when read with the figure.}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{(i) Since }\triangle ABC\text{ is equilateral,}
\displaystyle \angle CAB=\angle ABC=\angle ACB=60^\circ
\displaystyle \angle ADB=\angle ACB
\displaystyle \text{(Angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore \angle ADB=60^\circc31
\displaystyle \text{(ii) Since }ADBE\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADB+\angle AEB=180^\circ
\displaystyle 60^\circ+\angle AEB=180^\circ
\displaystyle \therefore \angle AEB=120^\circ
\displaystyle \text{Hence, (i) }\angle ADB=60^\circ\quad\text{and (ii) }\angle AEB=120^\circ.
\\

\displaystyle \textbf{Question 12: } \text{Given, }\angle CAB=75^\circ\text{ and }\angle CBA=50^\circ.
\displaystyle \text{Find the value of }\angle DAB+\angle ABD. c22\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete when read with the figure.}
\displaystyle \text{The final answer in the given solution is correct, but its working is incorrect.}
\displaystyle \text{It uses }5^\circ\text{ instead of }50^\circ\text{ and wrongly assumes an angle is }75^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ACB=180^\circ-\angle CAB-\angle CBA
\displaystyle \angle ACB=180^\circ-75^\circ-50^\circ=55^\circ
\displaystyle \angle ADB=\angle ACB
\displaystyle \text{(Angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore \angle ADB=55^\circ
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle DAB+\angle ABD+\angle ADB=180^\circ
\displaystyle \angle DAB+\angle ABD=180^\circ-55^\circ
\displaystyle \therefore \angle DAB+\angle ABD=125^\circ
\\

\displaystyle \textbf{Question 13: }ABCD\text{ is a cyclic quadrilateral in a circle with centre }O.
\displaystyle AB\text{ is a diameter. If }\angle ADC=130^\circ,\text{ find }\angle BAC. c23\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem is correct and complete when read with the figure.}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle \angle ABC+130^\circ=180^\circ
\displaystyle \therefore \angle ABC=50^\circ
\displaystyle \text{Since }AB\text{ is a diameter, }\angle ACB=90^\circ.
\displaystyle \text{(Angle in a semicircle)}
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle BAC+\angle ABC+\angle ACB=180^\circ
\displaystyle \angle BAC+50^\circ+90^\circ=180^\circ
\displaystyle \therefore \angle BAC=40^\circ
\\

\displaystyle \textbf{Question 14: } \text{In the given figure, }AOB\text{ is a diameter of the circle}
\displaystyle \text{and }\angle AOC=110^\circ.\text{ Find }\angle BDC. c24\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete.}
\displaystyle \text{The given solution is correct and complete, but it can be simplified.}
\displaystyle \text{Since }AOB\text{ is a diameter, }\angle AOB=180^\circ.
\displaystyle \angle AOC+\angle COB=180^\circ
\displaystyle 110^\circ+\angle COB=180^\circ
\displaystyle \therefore \angle COB=70^\circ
\displaystyle \text{The angle at the centre is twice the angle at the circumference}
\displaystyle \text{subtended by the same chord }BC.
\displaystyle \angle COB=2\angle BDC
\displaystyle 70^\circ=2\angle BDC
\displaystyle \therefore \angle BDC=35^\circ
\\

\displaystyle \textbf{Question 15: } \text{In the given figure, }O\text{ is the centre of the circle,}
\displaystyle \angle AOB=60^\circ\text{ and }\angle BDC=100^\circ.\text{ Find }\angle OBC. c25\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem is correct and complete when read with the figure.}
\displaystyle \text{The given solution is correct, but some reasons should be stated more clearly.}
\displaystyle \text{Let }\angle OBC=x.
\displaystyle \angle BCA=\frac{1}{2}\angle BOA
\displaystyle \text{(Angle at the circumference is half the angle at the centre}
\displaystyle \text{subtended by the same chord }BA\text{)}
\displaystyle \angle BCA=\frac{1}{2}\times60^\circ=30^\circ
\displaystyle \text{Since }A,D,C\text{ are collinear, }\angle BCD=\angle BCA=30^\circ.
\displaystyle \text{Since }O,D,B\text{ are collinear, }\angle CBD=\angle OBC=x.
\displaystyle \text{In }\triangle BCD,
\displaystyle \angle BCD+\angle BDC+\angle CBD=180^\circ
\displaystyle 30^\circ+100^\circ+x=180^\circ
\displaystyle x=50^\circ
\displaystyle \therefore \angle OBC=50^\circ
\\

\displaystyle \textbf{Question 16: }ABCD\text{ is a cyclic quadrilateral in which}
\displaystyle \angle DAC=27^\circ,\ \angle DBA=50^\circ\text{ and }\angle ADB=33^\circ.
\displaystyle \text{Calculate: (i) }\angle DBC\qquad\text{(ii) }\angle DCB\qquad\text{(iii) }\angle CAB. c26\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete when read with the figure.}
\displaystyle \text{The given solution is incorrect and incomplete.}
\displaystyle \text{It incorrectly states }\angle ACB=30^\circ\text{ instead of }33^\circ.
\displaystyle \text{It also does not present parts (i) and (ii) in the correct order.}
\displaystyle \text{(i) }\angle DBC=\angle DAC
\displaystyle \text{(Angles in the same segment subtended by chord }DC\text{)}
\displaystyle \therefore \angle DBC=27^\circ
\displaystyle \text{(ii) In }\triangle ADB,
\displaystyle \angle DAB=180^\circ-\angle DBA-\angle ADB
\displaystyle \angle DAB=180^\circ-50^\circ-33^\circ=97^\circ
\displaystyle \text{Opposite angles of a cyclic quadrilateral are supplementary.}
\displaystyle \angle DAB+\angle DCB=180^\circ
\displaystyle 97^\circ+\angle DCB=180^\circ
\displaystyle \therefore \angle DCB=83^\circ
\displaystyle \text{(iii) }\angle ACB=\angle ADB
\displaystyle \text{(Angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore \angle ACB=33^\circ
\displaystyle \angle ABC=\angle ABD+\angle DBC
\displaystyle \angle ABC=50^\circ+27^\circ=77^\circ
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle CAB=180^\circ-\angle ABC-\angle ACB
\displaystyle \angle CAB=180^\circ-77^\circ-33^\circ=70^\circ
\displaystyle \therefore \text{(i) }\angle DBC=27^\circ,\quad\text{(ii) }\angle DCB=83^\circ
\displaystyle \text{and (iii) }\angle CAB=70^\circ.
\\

\displaystyle \textbf{Question 17: } \text{In the given figure, }AB\text{ is a diameter of the circle}
\displaystyle \text{with centre }O.\text{ Also, }F,E,D\text{ are collinear and}
\displaystyle \angle ECD=\angle EDC=32^\circ.\text{ Prove that }\angle COF=\angle CEF. c27\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem is correct and complete when read with the figure.}
\displaystyle \text{The given solution has the correct conclusion but lacks some justification.}
\displaystyle \text{The statement }OF=OC\text{ is correct but unnecessary.}
\displaystyle \text{In }\triangle CED,
\displaystyle \angle ECD+\angle EDC+\angle CED=180^\circ
\displaystyle 32^\circ+32^\circ+\angle CED=180^\circ
\displaystyle \therefore \angle CED=116^\circ
\displaystyle \text{Since }F,E,D\text{ are collinear,}
\displaystyle \angle CEF+\angle CED=180^\circ
\displaystyle \angle CEF=180^\circ-116^\circ=64^\circ
\displaystyle \text{Also, }\angle FDC=\angle EDC=32^\circ
\displaystyle \angle FOC=2\angle FDC
\displaystyle \text{(Angle at the centre is twice the angle at the circumference}
\displaystyle \text{subtended by the same chord }FC\text{)}
\displaystyle \angle FOC=2\times32^\circ=64^\circ
\displaystyle \therefore \angle COF=\angle CEF=64^\circ
\\

\displaystyle \textbf{Question 18: } \text{In the given figure, }AB\text{ and }CD\text{ are straight lines}
\displaystyle \text{through the centre }O\text{ of the circle. If }\angle AOC=80^\circ
\displaystyle \text{and }\angle CDE=40^\circ,\text{ find: (i) }\angle DCE\qquad\text{(ii) }\angle ABC. c28\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem is correct and complete when read with the figure.}
\displaystyle \text{Part (i) of the given solution is correct, but part (ii) is incorrect.}
\displaystyle \text{The expression }180^\circ-100^\circ-50^\circ\text{ equals }30^\circ,\text{ not }40^\circ.
\displaystyle \text{(i) Since }CD\text{ is a diameter, }\angle CED=90^\circ.
\displaystyle \text{(Angle in a semicircle)}
\displaystyle \text{In }\triangle CDE,
\displaystyle \angle DCE+\angle CED+\angle CDE=180^\circ
\displaystyle \angle DCE+90^\circ+40^\circ=180^\circ
\displaystyle \therefore \angle DCE=50^\circ
\displaystyle \text{(ii) }\angle COE=2\angle CDE
\displaystyle \text{(Angle at the centre is twice the angle at the circumference}
\displaystyle \text{subtended by the same chord }CE\text{)}
\displaystyle \angle COE=2\times40^\circ=80^\circ
\displaystyle OC=OE\qquad\text{(Radii of the same circle)}
\displaystyle \therefore \angle OCE=\angle CEO=\frac{180^\circ-80^\circ}{2}=50^\circ
\displaystyle \text{Since }C,E,B\text{ are collinear, }\angle OCB=\angle OCE=50^\circ.
\displaystyle \text{Since }A,O,B\text{ are collinear,}
\displaystyle \angle COB=180^\circ-\angle COA=180^\circ-80^\circ=100^\circ
\displaystyle \text{In }\triangle OCB,
\displaystyle \angle OBC=180^\circ-\angle COB-\angle OCB
\displaystyle \angle OBC=180^\circ-100^\circ-50^\circ=30^\circ
\displaystyle \text{Since }A,O,B\text{ are collinear, }\angle ABC=\angle OBC.
\displaystyle \therefore \text{(i) }\angle DCE=50^\circ\quad\text{and (ii) }\angle ABC=30^\circ.
\\

\displaystyle \textbf{Question 19: } \text{In the given figure, }AC\text{ is a diameter of a circle}
\displaystyle \text{with centre }O.\text{ A circle is drawn on }AO\text{ as diameter.}
\displaystyle AE\text{ is a chord of the larger circle and intersects the smaller circle at }B. c29\displaystyle \text{Prove that }AB=BE.
\displaystyle \text{Answer:}
\displaystyle \text{Observations:}
\displaystyle \text{The problem statement is correct and complete when read with the figure.}
\displaystyle \text{The given solution is correct and complete.}
\displaystyle \text{The equality of the angles at }A\text{ should be justified by collinearity.}
\displaystyle \text{Since }AO\text{ is a diameter of the smaller circle,}
\displaystyle \angle ABO=90^\circ\qquad\text{(Angle in a semicircle)}
\displaystyle \text{Since }AC\text{ is a diameter of the larger circle,}
\displaystyle \angle AEC=90^\circ\qquad\text{(Angle in a semicircle)}
\displaystyle \text{Also, }A,O,C\text{ are collinear and }A,B,E\text{ are collinear.}
\displaystyle \therefore \angle OAB=\angle CAE
\displaystyle \therefore \triangle ABO\sim\triangle AEC\qquad\text{(AA similarity)}
\displaystyle \frac{AO}{AC}=\frac{AB}{AE}
\displaystyle \text{Since }AO\text{ is the radius and }AC\text{ is the diameter of the larger circle,}
\displaystyle \frac{AO}{AC}=\frac{1}{2}
\displaystyle \therefore \frac{AB}{AE}=\frac{1}{2}
\displaystyle AE=2AB
\displaystyle \text{Since }B\text{ lies between }A\text{ and }E,\ AE=AB+BE.
\displaystyle AB+BE=2AB
\displaystyle \therefore AB=BE
\\

\displaystyle \textbf{Question 20: }\text{In the following figure:}
\displaystyle \text{(i) If }\angle BAD=96^\circ,\text{ find }\angle BCD\text{ and }\angle BFE.
\displaystyle \text{(ii) Prove that }AD\parallel FE. c210\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle BAD+\angle BCD=180^\circ
\displaystyle 96^\circ+\angle BCD=180^\circ
\displaystyle \therefore \angle BCD=84^\circ
\displaystyle \text{Since }D,C\text{ and }E\text{ are collinear,}
\displaystyle \angle BCD+\angle BCE=180^\circ
\displaystyle 84^\circ+\angle BCE=180^\circ
\displaystyle \therefore \angle BCE=96^\circ
\displaystyle \text{Since }BCEF\text{ is a cyclic quadrilateral,}
\displaystyle \angle BCE+\angle BFE=180^\circ
\displaystyle 96^\circ+\angle BFE=180^\circ
\displaystyle \therefore \angle BFE=84^\circ
\displaystyle \therefore \angle BCD=84^\circ\text{ and }\angle BFE=84^\circ.
\displaystyle \text{(ii) Since }A,B\text{ and }F\text{ are collinear,}
\displaystyle \angle BAD+\angle BFE=96^\circ+84^\circ=180^\circ
\displaystyle \text{The co-interior angles are supplementary.}
\displaystyle \therefore AD\parallel FE.
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\displaystyle \textbf{Question 21: }\text{Prove that:}
\displaystyle \text{(i) A parallelogram inscribed in a circle is a rectangle.}
\displaystyle \text{(ii) A rhombus inscribed in a circle is a square.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }ABCD\text{ be a parallelogram inscribed in a circle.}
\displaystyle \angle BAD=\angle BCD\text{ (opposite angles of a parallelogram are equal)}
\displaystyle \angle BAD+\angle BCD=180^\circ\text{ (opposite angles of a cyclic quadrilateral are supplementary)}
\displaystyle \therefore \angle BAD=\angle BCD=90^\circ
\displaystyle \text{Similarly, }\angle ABC=\angle ADC=90^\circ
\displaystyle \therefore ABCD\text{ is a rectangle.}
\displaystyle \text{(ii) Let }ABCD\text{ be a rhombus inscribed in a circle.}
\displaystyle \text{A rhombus is a parallelogram.}
\displaystyle \therefore \angle BAD=\angle BCD=90^\circ\text{ and }\angle ABC=\angle ADC=90^\circ\text{ (from part (i))}
\displaystyle AB=BC=CD=DA\text{ (all sides of a rhombus are equal)}
\displaystyle \therefore ABCD\text{ is a square.}
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\displaystyle \textbf{Question 22: }\text{In the following figure, }AB=AC.\text{ Prove that }DEBC\text{ is an isosceles trapezium.} c45\displaystyle \text{Answer:}
\displaystyle AB=AC\text{ (given)}
\displaystyle \therefore \angle ABC=\angle ACB\text{ (angles opposite equal sides)}
\displaystyle \text{Since }BCED\text{ is a cyclic quadrilateral,}
\displaystyle \angle DBC+\angle CED=180^\circ
\displaystyle \text{Since }A,D,B\text{ are collinear, }\angle DBC=\angle ABC.
\displaystyle \therefore \angle ABC+\angle CED=180^\circ
\displaystyle \therefore \angle ACB+\angle CED=180^\circ
\displaystyle \therefore DE\parallel BC\text{ (co-interior angles are supplementary)}\qquad\ldots\text{(i)}
\displaystyle \text{Since }DE\parallel BC,\ \angle ADE=\angle ABC\text{ (corresponding angles)}
\displaystyle \text{Also, }\angle AED=\angle ACB\text{ (corresponding angles)}
\displaystyle \text{But }\angle ABC=\angle ACB.
\displaystyle \therefore \angle ADE=\angle AED
\displaystyle \therefore AD=AE\text{ (sides opposite equal angles)}
\displaystyle AB-AD=AC-AE
\displaystyle \therefore BD=CE\qquad\ldots\text{(ii)}
\displaystyle \text{From (i) and (ii), }DE\parallel BC\text{ and }BD=CE.
\displaystyle \therefore DEBC\text{ is an isosceles trapezium.}
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\displaystyle \textbf{Question 23: }\text{Two circles intersect at }P\text{ and }Q.\text{ Through }P,\text{ diameters }PA\text{ and }PB\text{ of the two circles are drawn.}
\displaystyle \text{Show that the points }A,Q\text{ and }B\text{ are collinear.} c46\displaystyle \text{Answer:}
\displaystyle \angle PQA=90^\circ\text{ (angle in a semicircle)}
\displaystyle \angle PQB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \angle AQB=\angle AQP+\angle PQB
\displaystyle \angle AQB=90^\circ+90^\circ=180^\circ
\displaystyle \therefore A,Q\text{ and }B\text{ are collinear.}
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\displaystyle \textbf{Question 24: }ABCD\text{ is a quadrilateral inscribed in a circle, with }\angle BAD=60^\circ.
\displaystyle O\text{ is the centre of the circle. Prove that:}
\displaystyle \angle OBD+\angle ODB=\angle CBD+\angle CDB. c47\displaystyle \text{Answer:}
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle BAD+\angle BCD=180^\circ
\displaystyle 60^\circ+\angle BCD=180^\circ
\displaystyle \therefore \angle BCD=120^\circ
\displaystyle \text{In }\triangle BCD,
\displaystyle \angle CBD+\angle CDB+\angle BCD=180^\circ
\displaystyle \therefore \angle CBD+\angle CDB=180^\circ-120^\circ=60^\circ\qquad\ldots\text{(i)}
\displaystyle \angle BOD=2\angle BAD\text{ (angle at the centre is twice the angle at the circumference)}
\displaystyle \therefore \angle BOD=2\times60^\circ=120^\circ
\displaystyle \text{In }\triangle BOD,
\displaystyle \angle OBD+\angle ODB+\angle BOD=180^\circ
\displaystyle \therefore \angle OBD+\angle ODB=180^\circ-120^\circ=60^\circ\qquad\ldots\text{(ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle \therefore \angle OBD+\angle ODB=\angle CBD+\angle CDB.
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\displaystyle \textbf{Question 25: }\text{The figure given below shows a circle with centre }O.
\displaystyle \text{Given }\angle AOC=a\text{ and }\angle ABC=b.
\displaystyle \text{(i) Find the relationship between }a\text{ and }b.
\displaystyle \text{(ii) Find }\angle OAB\text{ if }OABC\text{ is a parallelogram.} c44\displaystyle \text{Answer:}
\displaystyle \text{(i) }\angle AOC=a\text{ and }\angle ABC=b\text{ (given)}
\displaystyle \text{The angle subtended by the major arc }AC\text{ at the centre is }360^\circ-a.
\displaystyle \angle ABC=\frac{1}{2}(360^\circ-a)
\displaystyle b=\frac{1}{2}(360^\circ-a)
\displaystyle 2b=360^\circ-a
\displaystyle \therefore a+2b=360^\circ
\displaystyle \text{(ii) Since }OABC\text{ is a parallelogram,}
\displaystyle \angle AOC=\angle ABC\text{ (opposite angles of a parallelogram)}
\displaystyle \therefore a=b
\displaystyle a+2b=360^\circ
\displaystyle b+2b=360^\circ
\displaystyle 3b=360^\circ
\displaystyle \therefore b=120^\circ\text{ and }a=120^\circ
\displaystyle \angle OAB+\angle AOC=180^\circ\text{ (adjacent angles of a parallelogram)}
\displaystyle \angle OAB+120^\circ=180^\circ
\displaystyle \therefore \angle OAB=60^\circ
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\displaystyle \textbf{Question 26: }\text{Two chords }AB\text{ and }CD\text{ intersect at }P\text{ inside the circle.}
\displaystyle \text{Prove that the sum of the angles subtended by the arcs }AC\text{ and }BD
\displaystyle \text{at the centre }O\text{ is equal to twice }\angle APC. c48\displaystyle \text{Answer:}
\displaystyle \angle AOC=2\angle ADC\text{ (angle at the centre is twice the angle at the circumference)}
\displaystyle \text{Similarly, }\angle BOD=2\angle BAD
\displaystyle \text{Adding,}
\displaystyle \angle AOC+\angle BOD=2(\angle ADC+\angle BAD)\qquad\ldots\text{(i)}
\displaystyle \text{In }\triangle PAD,\ \angle APC\text{ is an exterior angle.}
\displaystyle \therefore \angle APC=\angle PAD+\angle ADP
\displaystyle \text{Since }A,P,B\text{ are collinear, }\angle PAD=\angle BAD.
\displaystyle \text{Since }C,P,D\text{ are collinear, }\angle ADP=\angle ADC.
\displaystyle \therefore \angle APC=\angle BAD+\angle ADC\qquad\ldots\text{(ii)}
\displaystyle \text{Using (i) and (ii),}
\displaystyle \angle AOC+\angle BOD=2(\angle BAD+\angle ADC)=2\angle APC
\displaystyle \therefore \angle AOC+\angle BOD=2\angle APC.
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\displaystyle \textbf{Question 27: }\text{In the given figure, }RS\text{ is a diameter of the circle, }NM\parallel RS
\displaystyle \text{and }\angle MRS=29^\circ.\text{ Find (i) }\angle RNM\text{ and (ii) }\angle NRM. c43\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }RS\text{ is a diameter,}
\displaystyle \angle RMS=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle RMS,
\displaystyle \angle RSM=180^\circ-\angle RMS-\angle MRS
\displaystyle \angle RSM=180^\circ-90^\circ-29^\circ=61^\circ
\displaystyle \text{Since }RNMS\text{ is a cyclic quadrilateral,}
\displaystyle \angle RNM+\angle RSM=180^\circ
\displaystyle \angle RNM+61^\circ=180^\circ
\displaystyle \therefore \angle RNM=119^\circ
\displaystyle \text{(ii) Since }NM\parallel RS,
\displaystyle \angle NMR=\angle MRS=29^\circ\text{ (alternate interior angles)}
\displaystyle \angle NMS=\angle NMR+\angle RMS c411.jpg
\displaystyle \angle NMS=29^\circ+90^\circ=119^\circ
\displaystyle \text{Since }RNMS\text{ is a cyclic quadrilateral,}
\displaystyle \angle NRS+\angle NMS=180^\circ
\displaystyle \angle NRS=180^\circ-119^\circ=61^\circ
\displaystyle \angle NRS=\angle NRM+\angle MRS
\displaystyle 61^\circ=\angle NRM+29^\circ
\displaystyle \therefore \angle NRM=32^\circ
\\

\displaystyle \textbf{Question 28: }\text{In the given figure, }AB\parallel CD\text{ and }O\text{ is the centre of the circle.}
\displaystyle \text{If }\angle ADC=25^\circ,\text{ find }\angle AEB.\text{ Give reasons.} c42\displaystyle \text{Answer:}
\displaystyle \text{Since }CD\text{ is a diameter,}
\displaystyle \angle CAD=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ACD=180^\circ-\angle CAD-\angle ADC
\displaystyle \angle ACD=180^\circ-90^\circ-25^\circ=65^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ACD+\angle ABD=180^\circ
\displaystyle \therefore \angle ABD=180^\circ-65^\circ=115^\circ
\displaystyle \text{Since }AB\parallel CD,
\displaystyle \angle DAB=\angle ADC=25^\circ\text{ (alternate interior angles)}
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle ADB=180^\circ-\angle DAB-\angle ABD
\displaystyle \angle ADB=180^\circ-25^\circ-115^\circ=40^\circ
\displaystyle \angle AEB=\angle ADB\text{ (angles in the same segment)}
\displaystyle \therefore \angle AEB=40^\circ.
\\

\displaystyle \textbf{Question 29: }\text{Two circles intersect at }P\text{ and }Q.\text{ Through }P,\text{ a straight line }APB
\displaystyle \text{is drawn to meet the circles at }A\text{ and }B.\text{ Through }Q,\text{ a straight line is drawn}
\displaystyle \text{to meet the circles at }C\text{ and }D.\text{ Prove that }AC\parallel BD. c41\displaystyle \text{Answer:}
\displaystyle \text{Since }A,P,Q,C\text{ are concyclic,}
\displaystyle \angle CAP=\angle CQP\text{ (angles in the same segment)}\qquad\ldots\text{(i)}
\displaystyle \text{Since }B,P,Q,D\text{ are concyclic,}
\displaystyle \angle PBD=\angle PQD\text{ (angles in the same segment)}\qquad\ldots\text{(ii)}
\displaystyle \text{Since }C,Q\text{ and }D\text{ are collinear,}
\displaystyle \angle CQP+\angle PQD=180^\circ
\displaystyle \text{Using (i) and (ii),}
\displaystyle \angle CAP+\angle PBD=180^\circ
\displaystyle \text{Therefore, the co-interior angles formed by the transversal }AB\text{ are supplementary.}
\displaystyle \therefore AC\parallel BD.
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\displaystyle \textbf{Question 30: }ABCD\text{ is a cyclic quadrilateral in which }AB\text{ and }DC,\text{ on being produced,}
\displaystyle \text{meet at }P\text{ such that }PA=PD.\text{ Prove that }AD\parallel BC. c45\displaystyle \text{Answer:}
\displaystyle PA=PD\text{ (given)}
\displaystyle \therefore \angle PAD=\angle PDA\text{ (angles opposite equal sides)}
\displaystyle \text{Since }P,D\text{ and }C\text{ are collinear,}
\displaystyle \angle CDA+\angle PDA=180^\circ
\displaystyle \therefore \angle CDA=180^\circ-\angle PDA
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle CDA=180^\circ
\displaystyle \angle ABC+180^\circ-\angle PDA=180^\circ
\displaystyle \therefore \angle ABC=\angle PDA=\angle PAD
\displaystyle \text{Since }P,A\text{ and }B\text{ are collinear,}
\displaystyle \angle DAB+\angle PAD=180^\circ
\displaystyle \therefore \angle DAB+\angle ABC=180^\circ
\displaystyle \text{Thus, the co-interior angles formed by the transversal }AB\text{ are supplementary.}
\displaystyle \therefore AD\parallel BC.
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\displaystyle \textbf{Question 31: }AB\text{ is a diameter of a circle, and }A,P,B\text{ and }R\text{ lie on the circle,}
\displaystyle \text{as shown in the figure. }A,P,Q\text{ and }R,B,Q\text{ are collinear. If }\angle PAB=35^\circ
\displaystyle \text{and }\angle PQB=25^\circ,\text{ find (i) }\angle PRB,\text{ (ii) }\angle PBR\text{ and (iii) }\angle BPR. c591\displaystyle \text{Answer:}
\displaystyle \text{(i) }\angle PRB=\angle PAB\text{ (angles in the same segment subtended by chord }PB\text{)}
\displaystyle \therefore \angle PRB=35^\circ
\displaystyle \text{(ii) Since }A,P\text{ and }Q\text{ are collinear, }\angle BAQ=\angle BAP=35^\circ.
\displaystyle \text{In }\triangle ABQ,
\displaystyle \angle ABQ=180^\circ-\angle BAQ-\angle AQB
\displaystyle \angle ABQ=180^\circ-35^\circ-25^\circ=120^\circ
\displaystyle \text{Since }R,B\text{ and }Q\text{ are collinear,}
\displaystyle \angle ABR=180^\circ-\angle ABQ=180^\circ-120^\circ=60^\circ
\displaystyle \angle APB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle APB,
\displaystyle \angle PBA=180^\circ-\angle APB-\angle PAB
\displaystyle \angle PBA=180^\circ-90^\circ-35^\circ=55^\circ
\displaystyle \angle PBR=\angle PBA+\angle ABR
\displaystyle \therefore \angle PBR=55^\circ+60^\circ=115^\circ
\displaystyle \text{(iii) In }\triangle PBR,
\displaystyle \angle BPR=180^\circ-\angle PRB-\angle PBR
\displaystyle \angle BPR=180^\circ-35^\circ-115^\circ
\displaystyle \therefore \angle BPR=30^\circ
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\displaystyle \textbf{Question 32: }\text{In the given figure, }SP\text{ bisects }\angle RPT\text{ and }PQRS\text{ is a cyclic quadrilateral.}
\displaystyle \text{Prove that }SQ=SR. c59\displaystyle \text{Answer:}
\displaystyle \text{Let }\angle SPR=\angle SPT=x.
\displaystyle \text{Since }Q,P\text{ and }T\text{ are collinear,}
\displaystyle \angle QPR+\angle RPT=180^\circ
\displaystyle \therefore \angle QPR=180^\circ-2x
\displaystyle \text{Since }PQRS\text{ is a cyclic quadrilateral,}
\displaystyle \angle QPR+\angle QSR=180^\circ
\displaystyle \therefore \angle QSR=2x
\displaystyle \text{Again, in }\triangle QRS,
\displaystyle \angle SQR+\angle SRQ=180^\circ-2x\qquad\ldots\text{(i)}
\displaystyle \text{Since }PQRS\text{ is a cyclic quadrilateral,}
\displaystyle \angle QPR+\angle SRQ=180^\circ
\displaystyle \therefore \angle SRQ=180^\circ-(180^\circ-2x)=2x?
\displaystyle \text{(Instead, }\angle SRQ=\angle SPR=x\text{ since }\angle SPR=\angle SQR\text{ by angles in the same segment.)}
\displaystyle \angle RQS=\angle RPS=x\text{ (angles in the same segment subtended by chord }RS\text{)}
\displaystyle \angle QRS=180^\circ-2x-x=x
\displaystyle \therefore \angle RQS=\angle QRS
\displaystyle \therefore SQ=SR\text{ (sides opposite equal angles in a triangle).}
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\displaystyle \textbf{Question 33: }\text{In the figure, }O\text{ is the centre of the circle, }\angle AOE=150^\circ
\displaystyle \text{and }\angle DAO=51^\circ.\text{ Calculate }\angle CEB\text{ and }\angle OCE. c58\displaystyle \text{Answer:}
\displaystyle \text{The reflex }\angle AOE=360^\circ-150^\circ=210^\circ
\displaystyle \angle ADE=\frac{1}{2}\times210^\circ=105^\circ
\displaystyle \text{(angle at the circumference is half the angle at the centre)}
\displaystyle \text{Since }A,O\text{ and }B\text{ are collinear,}
\displaystyle \angle DAB=\angle DAO=51^\circ
\displaystyle \text{Since }ABED\text{ is a cyclic quadrilateral,}
\displaystyle \angle DAB+\angle DEB=180^\circ
\displaystyle \angle DEB=180^\circ-51^\circ=129^\circ
\displaystyle \text{Since }D,E\text{ and }C\text{ are collinear,}
\displaystyle \angle CEB+\angle DEB=180^\circ
\displaystyle \therefore \angle CEB=180^\circ-129^\circ=51^\circ
\displaystyle \text{In }\triangle ADC,\ \angle DAC=\angle DAO=51^\circ
\displaystyle \text{and }\angle ADC=\angle ADE=105^\circ
\displaystyle \angle ACD=180^\circ-51^\circ-105^\circ=24^\circ
\displaystyle \text{Since }A,O,C\text{ are collinear and }D,E,C\text{ are collinear,}
\displaystyle \angle OCE=\angle ACD
\displaystyle \therefore \angle OCE=24^\circ
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\displaystyle \textbf{Question 34: }\text{In the given figure, }P\text{ and }Q\text{ are the centres of two circles intersecting}
\displaystyle \text{at }B\text{ and }C.\ A,C,D\text{ are collinear. Calculate the numerical value of }x. c57\displaystyle \text{Answer:}
\displaystyle \angle APB=150^\circ\text{ (given)}
\displaystyle \angle ACB=\frac{1}{2}\angle APB\text{ (angle at the circumference is half the angle at the centre)}
\displaystyle \therefore \angle ACB=\frac{1}{2}\times150^\circ=75^\circ
\displaystyle \text{Since }A,C\text{ and }D\text{ are collinear,}
\displaystyle \angle ACB+\angle BCD=180^\circ
\displaystyle \therefore \angle BCD=180^\circ-75^\circ=105^\circ
\displaystyle \angle BCD=\frac{1}{2}\text{ reflex }\angle BQD
\displaystyle 105^\circ=\frac{1}{2}(360^\circ-x)
\displaystyle 210^\circ=360^\circ-x
\displaystyle \therefore x=150^\circ
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\displaystyle \textbf{Question 35: }\text{In the figure given below, two circles intersect at }A\text{ and }B.
\displaystyle O\text{ is the centre of the smaller circle, and }A,O,B,C\text{ are concyclic.}
\displaystyle \text{Given }\angle APB=a,\text{ find in terms of }a\text{ the value of:}
\displaystyle \text{(i) the obtuse }\angle AOB,\quad\text{(ii) }\angle ACB,\quad\text{(iii) }\angle ADB.\text{ Give reasons.} c56\displaystyle \text{Answer:}
\displaystyle \text{(i) }\angle AOB=2\angle APB\text{ (angle at the centre is twice the angle at the circumference)}
\displaystyle \therefore \angle AOB=2a
\displaystyle \text{(ii) Since }AOBC\text{ is a cyclic quadrilateral,}
\displaystyle \angle AOB+\angle ACB=180^\circ
\displaystyle 2a+\angle ACB=180^\circ
\displaystyle \therefore \angle ACB=180^\circ-2a
\displaystyle \text{(iii) }\angle ADB=\angle ACB\text{ (angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore \angle ADB=180^\circ-2a
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\displaystyle \textbf{Question 36: }\text{In the given figure, }O\text{ is the centre of the circle and }\angle ABC=55^\circ.
\displaystyle \text{Calculate }x\text{ and }y. c55\displaystyle \text{Answer:}
\displaystyle \text{The angle at the centre is twice the angle at the circumference subtended by the same arc }AC.
\displaystyle \angle AOC=2\angle ABC
\displaystyle x=2\times55^\circ
\displaystyle \therefore x=110^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADC+\angle ABC=180^\circ
\displaystyle y+55^\circ=180^\circ
\displaystyle \therefore y=125^\circ
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\displaystyle \textbf{Question 37: }\text{In the given figure, }A\text{ is the centre of the circle, }ABCD\text{ is a parallelogram}
\displaystyle \text{and }C,D,E\text{ are collinear. Prove that }\angle BCD=2\angle ABE. c54\displaystyle \text{Answer:}
\displaystyle \angle BAD=2\angle BED
\displaystyle \text{(angle at the centre is twice the angle at the circumference subtended by the same chord }BD\text{)}
\displaystyle \text{Since }ABCD\text{ is a parallelogram, }AB\parallel CD.
\displaystyle \text{Also, }C,D\text{ and }E\text{ are collinear.}
\displaystyle \therefore AB\parallel DE
\displaystyle \therefore \angle BED=\angle ABE\text{ (alternate interior angles)}
\displaystyle \therefore \angle BAD=2\angle ABE\qquad\ldots\text{(i)}
\displaystyle \angle BAD=\angle BCD\text{ (opposite angles of a parallelogram are equal)}
\displaystyle \text{Using (i),}
\displaystyle \therefore \angle BCD=2\angle ABE.
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\displaystyle \textbf{Question 38: }ABCD\text{ is a cyclic quadrilateral in which }AB\parallel DC\text{ and }AB
\displaystyle \text{is a diameter of the circle. Given }\angle BED=65^\circ,\text{ calculate:}
\displaystyle \text{(i) }\angle DAB\qquad\text{(ii) }\angle BDC. c53\displaystyle \text{Answer:}
\displaystyle \text{(i) }\angle DAB=\angle DEB\text{ (angles in the same segment subtended by chord }DB\text{)}
\displaystyle \therefore \angle DAB=65^\circ
\displaystyle \text{(ii) Since }AB\text{ is a diameter,}
\displaystyle \angle ADB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle DBA=180^\circ-\angle DAB-\angle ADB
\displaystyle \angle DBA=180^\circ-65^\circ-90^\circ=25^\circ
\displaystyle \text{Since }AB\parallel DC,
\displaystyle \angle BDC=\angle DBA\text{ (alternate interior angles)}
\displaystyle \therefore \angle BDC=25^\circ
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\displaystyle \textbf{Question 39: }\text{In the given figure, }AB\text{ is the diameter of the circle. Chord }ED\parallel AB
\displaystyle \text{and }\angle EAB=63^\circ.\text{ Calculate (i) }\angle EBA\text{ and (ii) }\angle BCD. c52\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AB\text{ is a diameter,}
\displaystyle \angle AEB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle AEB,
\displaystyle \angle EBA=180^\circ-\angle AEB-\angle EAB
\displaystyle \angle EBA=180^\circ-90^\circ-63^\circ
\displaystyle \therefore \angle EBA=27^\circ
\displaystyle \text{(ii) Since }ED\parallel AB,
\displaystyle \angle DEB=\angle EBA=27^\circ\text{ (alternate interior angles)}
\displaystyle \text{Since }EBCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle DEB+\angle DCB=180^\circ
\displaystyle 27^\circ+\angle DCB=180^\circ
\displaystyle \therefore \angle DCB=153^\circ
\displaystyle \therefore \angle BCD=153^\circ
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\displaystyle \textbf{Question 40: }\text{The sides }AB\text{ and }CD\text{ of a cyclic quadrilateral }ABCD\text{ are produced}
\displaystyle \text{to meet at }E,\text{ and the sides }DA\text{ and }CB\text{ are produced to meet at }F.
\displaystyle \text{If }\angle BEC=42^\circ\text{ and }\angle BAD=98^\circ,\text{ find:}
\displaystyle \text{(i) }\angle AFB\qquad\text{(ii) }\angle ADC. c51.jpg\displaystyle \text{Answer:}
\displaystyle \text{Since }F,A\text{ and }D\text{ are collinear,}
\displaystyle \angle BAF+\angle BAD=180^\circ
\displaystyle \angle BAF=180^\circ-98^\circ=82^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle BAD+\angle BCD=180^\circ
\displaystyle \angle BCD=180^\circ-98^\circ=82^\circ
\displaystyle \text{Since }D,C\text{ and }E\text{ are collinear,}
\displaystyle \angle BCD+\angle BCE=180^\circ
\displaystyle \angle BCE=180^\circ-82^\circ=98^\circ
\displaystyle \text{In }\triangle BCE,
\displaystyle \angle CBE=180^\circ-\angle BCE-\angle BEC
\displaystyle \angle CBE=180^\circ-98^\circ-42^\circ=40^\circ
\displaystyle \text{(i) }\angle ABF=\angle CBE=40^\circ\text{ (vertically opposite angles)}
\displaystyle \text{In }\triangle ABF,
\displaystyle \angle AFB=180^\circ-\angle BAF-\angle ABF
\displaystyle \angle AFB=180^\circ-82^\circ-40^\circ
\displaystyle \therefore \angle AFB=58^\circ
\displaystyle \text{(ii) Since }A,B\text{ and }E\text{ are collinear,}
\displaystyle \angle ABC+\angle CBE=180^\circ
\displaystyle \angle ABC=180^\circ-40^\circ=140^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle \angle ADC=180^\circ-140^\circ
\displaystyle \therefore \angle ADC=40^\circ
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\displaystyle \textbf{Question 41: }\text{In the given figure, }AB\text{ is the diameter of the circle with centre }O.
\displaystyle DO\parallel CB\text{ and }\angle DCB=120^\circ.\text{ Calculate:}
\displaystyle \text{(i) }\angle DAB\quad\text{(ii) }\angle DBA\quad\text{(iii) }\angle DBC\quad\text{(iv) }\angle ADC.
\displaystyle \text{Also, prove that }\triangle AOD\text{ is an equilateral triangle.} c61\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle DAB+\angle DCB=180^\circ
\displaystyle \angle DAB+120^\circ=180^\circ
\displaystyle \therefore \angle DAB=60^\circ
\displaystyle \text{(ii) Since }AB\text{ is a diameter,}
\displaystyle \angle ADB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle DBA=180^\circ-\angle DAB-\angle ADB
\displaystyle \angle DBA=180^\circ-60^\circ-90^\circ
\displaystyle \therefore \angle DBA=30^\circ
\displaystyle \text{(iii) Since }O\text{ lies on }AB,\ \angle DBO=\angle DBA=30^\circ.
\displaystyle OD=OB\text{ (radii of the same circle)}
\displaystyle \therefore \angle ODB=\angle DBO=30^\circ\text{ (angles opposite equal sides)}
\displaystyle \text{Since }DO\parallel CB,
\displaystyle \angle DBC=\angle ODB\text{ (alternate interior angles)}
\displaystyle \therefore \angle DBC=30^\circ
\displaystyle \text{(iv) }\angle ABC=\angle DBA+\angle DBC
\displaystyle \angle ABC=30^\circ+30^\circ=60^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADC+\angle ABC=180^\circ
\displaystyle \angle ADC=180^\circ-60^\circ
\displaystyle \therefore \angle ADC=120^\circ
\displaystyle \text{To prove that }\triangle AOD\text{ is equilateral:}
\displaystyle OA=OD\text{ (radii of the same circle)}
\displaystyle \therefore \angle OAD=\angle ADO\text{ (angles opposite equal sides)}
\displaystyle \text{Since }A,O\text{ and }B\text{ are collinear, }\angle OAD=\angle DAB=60^\circ.
\displaystyle \therefore \angle ADO=60^\circ
\displaystyle \angle AOD=180^\circ-60^\circ-60^\circ=60^\circ
\displaystyle \therefore \angle OAD=\angle ADO=\angle AOD=60^\circ
\displaystyle \therefore \triangle AOD\text{ is an equilateral triangle.}
\\

\displaystyle \textbf{Question 42: }\text{In the given figure, }I\text{ is the incentre of }\triangle ABC.\ BI,\text{ when produced,}
\displaystyle \text{meets the circumference of the circle at }D.\text{ Given }\angle BAC=55^\circ\text{ and }\angle ACB=65^\circ,
\displaystyle \text{calculate: (i) }\angle DCA\quad\text{(ii) }\angle DAC\quad\text{(iii) }\angle DCI\quad\text{(iv) }\angle AIC. c62\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ABC=180^\circ-\angle BAC-\angle ACB
\displaystyle \angle ABC=180^\circ-55^\circ-65^\circ=60^\circ
\displaystyle \text{Since }BI\text{ bisects }\angle ABC\text{ and }B,I,D\text{ are collinear,}
\displaystyle \angle ABD=\angle DBC=\frac{1}{2}\angle ABC=30^\circ
\displaystyle \text{(i) }\angle DCA=\angle DBA\text{ (angles in the same segment subtended by chord }DA\text{)}
\displaystyle \therefore \angle DCA=30^\circ
\displaystyle \text{(ii) }\angle DAC=\angle DBC\text{ (angles in the same segment subtended by chord }DC\text{)}
\displaystyle \therefore \angle DAC=30^\circ
\displaystyle \text{(iii) Since }CI\text{ bisects }\angle ACB,
\displaystyle \angle ACI=\frac{1}{2}\angle ACB=\frac{65^\circ}{2}=32.5^\circ
\displaystyle \angle DCI=\angle DCA+\angle ACI
\displaystyle \angle DCI=30^\circ+32.5^\circ
\displaystyle \therefore \angle DCI=62.5^\circ
\displaystyle \text{(iv) Since }AI\text{ bisects }\angle BAC,
\displaystyle \angle IAC=\frac{1}{2}\angle BAC=\frac{55^\circ}{2}=27.5^\circ
\displaystyle \text{In }\triangle AIC,
\displaystyle \angle AIC=180^\circ-\angle IAC-\angle ACI
\displaystyle \angle AIC=180^\circ-27.5^\circ-32.5^\circ
\displaystyle \therefore \angle AIC=120^\circ
\\

\displaystyle \textbf{Question 43: }\triangle ABC\text{ is inscribed in a circle. The bisectors of }\angle BAC,\angle ABC
\displaystyle \text{and }\angle ACB\text{ meet the circumference at }P,Q\text{ and }R,\text{ respectively. Prove that:}
\displaystyle \text{(i) }\angle ABC=2\angle APQ
\displaystyle \text{(ii) }\angle ACB=2\angle APR
\displaystyle \text{(iii) }\angle QPR=90^\circ-\frac{1}{2}\angle BAC. c63\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }BQ\text{ bisects }\angle ABC,
\displaystyle \angle ABQ=\frac{1}{2}\angle ABC
\displaystyle \angle APQ=\angle ABQ\text{ (angles in the same segment subtended by chord }AQ\text{)}
\displaystyle \therefore \angle ABC=2\angle ABQ=2\angle APQ
\displaystyle \therefore \angle ABC=2\angle APQ.
\displaystyle \text{(ii) Since }CR\text{ bisects }\angle ACB,
\displaystyle \angle ACR=\frac{1}{2}\angle ACB
\displaystyle \angle APR=\angle ACR\text{ (angles in the same segment subtended by chord }AR\text{)}
\displaystyle \therefore \angle ACB=2\angle ACR=2\angle APR
\displaystyle \therefore \angle ACB=2\angle APR.
\displaystyle \text{(iii) Adding the results obtained in (i) and (ii),}
\displaystyle \angle ABC+\angle ACB=2\angle APQ+2\angle APR
\displaystyle \angle ABC+\angle ACB=2(\angle APQ+\angle APR)
\displaystyle \text{Since the ray }PA\text{ lies inside }\angle QPR,
\displaystyle \angle QPR=\angle QPA+\angle APR
\displaystyle \text{But }\angle QPA=\angle APQ.
\displaystyle \therefore \angle QPR=\angle APQ+\angle APR
\displaystyle \therefore \angle ABC+\angle ACB=2\angle QPR
\displaystyle \text{In }\triangle ABC,\ \angle ABC+\angle ACB=180^\circ-\angle BAC
\displaystyle \therefore 180^\circ-\angle BAC=2\angle QPR
\displaystyle \therefore \angle QPR=90^\circ-\frac{1}{2}\angle BAC.
\\

\displaystyle \textbf{Question 44: }\text{Calculate the angles }x,y\text{ and }z\text{ if }\frac{x}{3}=\frac{y}{4}=\frac{z}{5}. c64\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k.
\displaystyle \therefore x=3k,\quad y=4k,\quad z=5k
\displaystyle \text{In }\triangle PBC,
\displaystyle \angle BPC=y=4k
\displaystyle \angle PCB=x=3k\text{ (vertically opposite angles)}
\displaystyle \therefore \angle PBC=180^\circ-4k-3k=180^\circ-7k
\displaystyle \text{Since }A,B,P\text{ are collinear,}
\displaystyle \angle ABC=180^\circ-\angle PBC=7k
\displaystyle \text{Similarly, in }\triangle QCD,
\displaystyle \angle DQC=z=5k
\displaystyle \angle QCD=x=3k\text{ (vertically opposite angles)}
\displaystyle \therefore \angle QDC=180^\circ-5k-3k=180^\circ-8k
\displaystyle \text{Since }A,D,Q\text{ are collinear,}
\displaystyle \angle ADC=180^\circ-\angle QDC=8k
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle 7k+8k=180^\circ
\displaystyle \therefore k=12^\circ
\displaystyle x=3k=36^\circ,\qquad y=4k=48^\circ,\qquad z=5k=60^\circ
\displaystyle \therefore x=36^\circ,\ y=48^\circ,\ z=60^\circ.
\\

\displaystyle \textbf{Question 45: }\text{In the given figure, }AB=AC=CD\text{ and }\angle ADC=38^\circ.\text{ Calculate:}
\displaystyle \text{(i) }\angle ABC\qquad\text{(ii) }\angle BEC.\qquad\text{[1995]} c65\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AC=CD,
\displaystyle \angle CAD=\angle ADC=38^\circ\text{ (angles opposite equal sides)}
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ACD=180^\circ-38^\circ-38^\circ=104^\circ
\displaystyle \text{Since }B,C\text{ and }D\text{ are collinear,}
\displaystyle \angle ACB+\angle ACD=180^\circ
\displaystyle \angle ACB=180^\circ-104^\circ=76^\circ
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB\text{ (angles opposite equal sides)}
\displaystyle \therefore \angle ABC=76^\circ
\displaystyle \text{(ii) In }\triangle ABC,
\displaystyle \angle BAC=180^\circ-\angle ABC-\angle ACB
\displaystyle \angle BAC=180^\circ-76^\circ-76^\circ=28^\circ
\displaystyle \angle BEC=\angle BAC\text{ (angles in the same segment subtended by chord }BC\text{)}
\displaystyle \therefore \angle BEC=28^\circ
\\

\displaystyle \textbf{Question 46: }\text{In the given figure, }AC\text{ is the diameter of the circle with centre }O.
\displaystyle \text{Chord }BD\perp AC.\text{ Write down the angles }p,q\text{ and }r\text{ in terms of }x.\text{ [1996]} c66\displaystyle \text{Answer:}
\displaystyle \angle AOB=2\angle ACB\text{ (angle at the centre is twice the angle at the circumference subtended by chord }AB\text{)}
\displaystyle x=2q
\displaystyle \therefore q=\frac{x}{2}
\displaystyle \angle ADB=\angle ACB=\frac{x}{2}\text{ (angles in the same segment subtended by chord }AB\text{)}
\displaystyle \text{Since }AC\text{ is a diameter,}
\displaystyle \angle ADC=90^\circ\text{ (angle in a semicircle)}
\displaystyle \angle ADB+\angle BDC=90^\circ
\displaystyle \frac{x}{2}+r=90^\circ
\displaystyle \therefore r=90^\circ-\frac{x}{2}
\displaystyle \text{Also, }\angle ABC=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ABC,
\displaystyle p+q=90^\circ
\displaystyle p=90^\circ-q
\displaystyle \therefore p=90^\circ-\frac{x}{2}
\displaystyle \therefore p=90^\circ-\frac{x}{2},\qquad q=\frac{x}{2},\qquad r=90^\circ-\frac{x}{2}.
\\

\displaystyle \textbf{Question 47: }\text{In the given figure, }AC\text{ is the diameter of the circle with centre }O.
\displaystyle CD\parallel BE,\ \angle AOB=80^\circ\text{ and }\angle ACE=10^\circ.\text{ Calculate:}
\displaystyle \text{(i) }\angle BEC\qquad\text{(ii) }\angle BCD\qquad\text{(iii) }\angle CED.\qquad\text{[1998]} c67\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }AC\text{ is a diameter, }\angle AOC=180^\circ.
\displaystyle \angle BOC=180^\circ-\angle AOB
\displaystyle \angle BOC=180^\circ-80^\circ=100^\circ
\displaystyle \angle BOC=2\angle BEC\text{ (angle at the centre is twice the angle at the circumference)}
\displaystyle 100^\circ=2\angle BEC
\displaystyle \therefore \angle BEC=50^\circ
\displaystyle \text{(ii) Since }CD\parallel BE,
\displaystyle \angle DCE=\angle BEC=50^\circ\text{ (alternate interior angles)}
\displaystyle \angle ACB=\frac{1}{2}\angle AOB\text{ (angle at the circumference is half the angle at the centre)}
\displaystyle \angle ACB=\frac{1}{2}\times80^\circ=40^\circ
\displaystyle \angle BCD=\angle BCA+\angle ACE+\angle ECD
\displaystyle \angle BCD=40^\circ+10^\circ+50^\circ
\displaystyle \therefore \angle BCD=100^\circ
\displaystyle \text{(iii) Since }BCDE\text{ is a cyclic quadrilateral,}
\displaystyle \angle BCD+\angle BED=180^\circ
\displaystyle \angle BED=180^\circ-100^\circ=80^\circ
\displaystyle \angle BED=\angle BEC+\angle CED
\displaystyle 80^\circ=50^\circ+\angle CED
\displaystyle \therefore \angle CED=30^\circ
\\

\displaystyle \textbf{Question 48: }\text{In the given figure, }AE\text{ is the diameter of the circle. Write down the numerical value of}
\displaystyle \angle ABC+\angle CDE.\text{ Give reasons for your answer. }\text{[1998]} c68\displaystyle \text{Answer:}
\displaystyle \text{Since }ABCE\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle AEC=180^\circ\qquad\ldots\text{(i)}
\displaystyle \text{Since }ACDE\text{ is a cyclic quadrilateral,}
\displaystyle \angle CDE+\angle CAE=180^\circ\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle \angle ABC+\angle CDE+\angle AEC+\angle CAE=360^\circ
\displaystyle \text{Since }AE\text{ is a diameter,}
\displaystyle \angle ACE=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ACE,
\displaystyle \angle AEC+\angle CAE=180^\circ-90^\circ=90^\circ
\displaystyle \therefore \angle ABC+\angle CDE=360^\circ-90^\circ
\displaystyle \therefore \angle ABC+\angle CDE=270^\circ
\\

\displaystyle \textbf{Question 49: }\text{In the given figure, }AC\text{ is the diameter and }AC\parallel ED.
\displaystyle \text{If }\angle CBE=64^\circ,\text{ calculate }\angle DEC.\text{ [1991]} c69.jpg\displaystyle \text{Answer:}
\displaystyle \text{Since }AC\text{ is a diameter,}
\displaystyle \angle ABC=90^\circ\text{ (angle in a semicircle)}
\displaystyle \angle ABC=\angle ABE+\angle EBC
\displaystyle 90^\circ=\angle ABE+64^\circ
\displaystyle \therefore \angle ABE=26^\circ
\displaystyle \angle ACE=\angle ABE\text{ (angles in the same segment subtended by chord }AE\text{)}
\displaystyle \therefore \angle ACE=26^\circ
\displaystyle \text{Since }AC\parallel ED,
\displaystyle \angle DEC=\angle ACE\text{ (alternate interior angles)}
\displaystyle \therefore \angle DEC=26^\circ
\\

\displaystyle \textbf{Question 50: }\text{Use the given figure to find: (i) }\angle BAD\qquad\text{(ii) }\angle DQB.\text{ [1987]} c610\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }A,B,P\text{ are collinear and }D,C,P\text{ are collinear,}
\displaystyle \angle APD=40^\circ\text{ and }\angle ADP=85^\circ
\displaystyle \text{In }\triangle ADP,
\displaystyle \angle BAD=180^\circ-\angle ADP-\angle APD
\displaystyle \angle BAD=180^\circ-85^\circ-40^\circ
\displaystyle \therefore \angle BAD=55^\circ
\displaystyle \text{(ii) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle \angle ABC=180^\circ-85^\circ=95^\circ
\displaystyle \text{Since }A,D,Q\text{ are collinear and }B,C,Q\text{ are collinear,}
\displaystyle \angle QAB=\angle DAB=55^\circ
\displaystyle \angle ABQ=\angle ABC=95^\circ
\displaystyle \text{In }\triangle ABQ,
\displaystyle \angle AQB=180^\circ-\angle QAB-\angle ABQ
\displaystyle \angle AQB=180^\circ-55^\circ-95^\circ=30^\circ
\displaystyle \text{Since }A,D,Q\text{ are collinear, }\angle DQB=\angle AQB.
\displaystyle \therefore \angle DQB=30^\circ
\\

\displaystyle \textbf{Question 51: }\text{In the given figure, }AB\text{ is the diameter of the circle with centre }O\text{ and }DC\parallel AB.
\displaystyle \text{If }\angle CAB=x,\text{ find in terms of }x\text{: (i) }\angle COB\quad\text{(ii) }\angle DOC
\displaystyle \text{(iii) }\angle DAC\quad\text{(iv) }\angle ADC.\qquad\text{[1991]} c79\displaystyle \text{Answer:}
\displaystyle \text{(i) }\angle COB=2\angle CAB
\displaystyle \text{(angle at the centre is twice the angle at the circumference subtended by chord }CB\text{)}
\displaystyle \therefore \angle COB=2x
\displaystyle \text{(ii) Since }DC\parallel AB\text{ and }A,O,B\text{ are collinear, }DC\parallel OB.
\displaystyle \therefore \angle OCD=\angle COB=2x\text{ (alternate interior angles)}
\displaystyle OC=OD\text{ (radii of the same circle)}
\displaystyle \therefore \angle ODC=\angle OCD=2x\text{ (angles opposite equal sides)}
\displaystyle \text{In }\triangle OCD,
\displaystyle \angle DOC=180^\circ-\angle OCD-\angle ODC
\displaystyle \therefore \angle DOC=180^\circ-2x-2x=180^\circ-4x
\displaystyle \text{(iii) }\angle DAC=\frac{1}{2}\angle DOC
\displaystyle \text{(angle at the circumference is half the angle at the centre subtended by chord }DC\text{)}
\displaystyle \angle DAC=\frac{1}{2}(180^\circ-4x)
\displaystyle \therefore \angle DAC=90^\circ-2x
\displaystyle \text{(iv) Since }DC\parallel AB,
\displaystyle \angle ACD=\angle CAB=x\text{ (alternate interior angles)}
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ADC=180^\circ-\angle DAC-\angle ACD
\displaystyle \angle ADC=180^\circ-(90^\circ-2x)-x
\displaystyle \therefore \angle ADC=90^\circ+x
\\

\displaystyle \textbf{Question 52: }\text{In the figure, }AB\text{ is the diameter of the circle with centre }O.
\displaystyle \text{If }\angle BCD=130^\circ,\text{ find (i) }\angle DAB\text{ and (ii) }\angle DBA.\text{ [2012]} c78\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle DAB+\angle BCD=180^\circ
\displaystyle \angle DAB=180^\circ-130^\circ
\displaystyle \therefore \angle DAB=50^\circ
\displaystyle \text{(ii) Since }AB\text{ is a diameter,}
\displaystyle \angle ADB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ADB,
\displaystyle \angle DAB+\angle ADB+\angle DBA=180^\circ
\displaystyle 50^\circ+90^\circ+\angle DBA=180^\circ
\displaystyle \therefore \angle DBA=40^\circ
\\

\displaystyle \textbf{Question 53: }\text{In the given figure, }PQ\text{ is the diameter of the circle with centre }O.
\displaystyle \text{Given }\angle ROS=42^\circ,\text{ calculate }\angle RTS.\text{ [1992]} c77\displaystyle \text{Answer:}
\displaystyle \text{Join }P\text{ and }S.
\displaystyle \angle RPS=\frac{1}{2}\angle ROS
\displaystyle \text{(angle at the circumference is half the angle at the centre subtended by chord }RS\text{)}
\displaystyle \angle RPS=\frac{1}{2}\times42^\circ=21^\circc715
\displaystyle \text{Since }P,R\text{ and }T\text{ are collinear,}
\displaystyle \angle SPT=\angle SPR=21^\circ
\displaystyle \text{Since }PQ\text{ is a diameter,}
\displaystyle \angle PSQ=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{Since }Q,S\text{ and }T\text{ are collinear,}
\displaystyle \angle PST=180^\circ-\angle PSQ=90^\circ
\displaystyle \text{In }\triangle PST,
\displaystyle \angle PTS=180^\circ-\angle SPT-\angle PST
\displaystyle \angle PTS=180^\circ-21^\circ-90^\circ=69^\circ
\displaystyle \text{Since }P,R\text{ and }T\text{ are collinear, }\angle RTS=\angle PTS.
\displaystyle \therefore \angle RTS=69^\circ
\\

\displaystyle \textbf{Question 54: }\text{In the given figure, }PQ\text{ is the diameter of the circle and chord }SR\parallel PQ.
\displaystyle \text{Given }\angle PQR=58^\circ,\text{ calculate (i) }\angle RPQ\text{ and (ii) }\angle STP.\text{ [1989]} c76\displaystyle \text{Answer:}
\displaystyle \text{Join }P\text{ and }R.
\displaystyle \text{(i) Since }PQ\text{ is a diameter,}
\displaystyle \angle PRQ=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle PQR,c714
\displaystyle \angle RPQ=180^\circ-\angle PRQ-\angle PQR
\displaystyle \angle RPQ=180^\circ-90^\circ-58^\circ
\displaystyle \therefore \angle RPQ=32^\circ
\displaystyle \text{(ii) Since }SR\parallel PQ,
\displaystyle \angle PSR=\angle RPQ=32^\circ\text{ (alternate interior angles)}
\displaystyle \text{Since }P,T,S,R\text{ are concyclic,}
\displaystyle \angle PSR+\angle PTR=180^\circ
\displaystyle \angle PTR=180^\circ-32^\circ=148^\circ
\displaystyle \text{Since }S,T,P\text{ determine the same angle at }T,\ \angle STP=\angle PTR.
\displaystyle \therefore \angle STP=148^\circ
\\

\displaystyle \textbf{Question 55: }AB\text{ is the diameter of the circle with centre }O.\ OD\parallel BC\text{ and }\angle AOD=60^\circ.
\displaystyle \text{Calculate the numerical values of: (i) }\angle ABD\quad\text{(ii) }\angle DBC\quad\text{(iii) }\angle ADC.\text{ [1987]} c75\displaystyle \text{Answer:}
\displaystyle \text{Join }B\text{ and }D.
\displaystyle \text{(i) }\angle ABD=\frac{1}{2}\angle AOD
\displaystyle \text{(angle at the circumference is half the angle at the centre subtended by chord }AD\text{)}
\displaystyle \angle ABD=\frac{1}{2}\times60^\circ
\displaystyle \therefore \angle ABD=30^\circ
\displaystyle \text{(ii) Since }AB\text{ is a diameter,}
\displaystyle \angle ADB=90^\circ\text{ (angle in a semicircle)}
\displaystyle OA=OD\text{ (radii of the same circle)}
\displaystyle \therefore \angle OAD=\angle ADO
\displaystyle \text{In }\triangle AOD,c713
\displaystyle \angle OAD+\angle ADO+\angle AOD=180^\circ
\displaystyle 2\angle ADO+60^\circ=180^\circ
\displaystyle \therefore \angle ADO=60^\circ
\displaystyle \angle ODB=\angle ADB-\angle ADO
\displaystyle \angle ODB=90^\circ-60^\circ=30^\circ
\displaystyle \text{Since }OD\parallel BC,
\displaystyle \angle DBC=\angle ODB\text{ (alternate interior angles)}
\displaystyle \therefore \angle DBC=30^\circ
\displaystyle \text{(iii) }\angle ABC=\angle ABD+\angle DBC
\displaystyle \angle ABC=30^\circ+30^\circ=60^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle \angle ADC=180^\circ-60^\circ
\displaystyle \therefore \angle ADC=120^\circ
\\

\displaystyle \textbf{Question 56: }\text{In the given figure, the centre }O\text{ of the smaller circle lies on the circumference}
\displaystyle \text{of the bigger circle. If }\angle APB=75^\circ\text{ and }\angle BCD=40^\circ,\text{ find:}
\displaystyle \text{(i) }\angle AOB\quad\text{(ii) }\angle ACB\quad\text{(iii) }\angle ABD\quad\text{(iv) }\angle ADB.\text{ [1984]} c74\displaystyle \text{Answer:}
\displaystyle \text{Join }A\text{ and }B.
\displaystyle \text{(i) }\angle AOB=2\angle APB
\displaystyle \text{(angle at the centre is twice the angle at the circumference subtended by chord }AB\text{)}
\displaystyle \angle AOB=2\times75^\circ
\displaystyle \therefore \angle AOB=150^\circ
\displaystyle \text{(ii) Since }AOBC\text{ is a cyclic quadrilateral,}
\displaystyle \angle AOB+\angle ACB=180^\circc712
\displaystyle \angle ACB=180^\circ-150^\circ
\displaystyle \therefore \angle ACB=30^\circ
\displaystyle \text{(iii) }\angle ACD=\angle ACB+\angle BCD
\displaystyle \angle ACD=30^\circ+40^\circ=70^\circ
\displaystyle \text{Since }ABDC\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABD+\angle ACD=180^\circ
\displaystyle \angle ABD=180^\circ-70^\circ
\displaystyle \therefore \angle ABD=110^\circ
\displaystyle \text{(iv) Since }ADBO\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADB+\angle AOB=180^\circ
\displaystyle \angle ADB=180^\circ-150^\circ
\displaystyle \therefore \angle ADB=30^\circ
\\

\displaystyle \textbf{Question 57: }\text{In the figure, }\angle BAD=65^\circ,\ \angle ABD=70^\circ\text{ and }\angle BDC=45^\circ.
\displaystyle \text{Find (i) }\angle BCD\text{ and (ii) }\angle ACB.\text{ Hence, show that }AC\text{ is the diameter.} c73\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle BAD+\angle BCD=180^\circ
\displaystyle \angle BCD=180^\circ-65^\circ
\displaystyle \therefore \angle BCD=115^\circ
\displaystyle \text{(ii) In }\triangle ABD,
\displaystyle \angle ADB=180^\circ-\angle BAD-\angle ABD
\displaystyle \angle ADB=180^\circ-65^\circ-70^\circ
\displaystyle \therefore \angle ADB=45^\circ
\displaystyle \angle ACB=\angle ADB\text{ (angles in the same segment subtended by chord }AB\text{)}
\displaystyle \therefore \angle ACB=45^\circ
\displaystyle \angle ADC=\angle ADB+\angle BDC
\displaystyle \angle ADC=45^\circ+45^\circ=90^\circ
\displaystyle \text{Since chord }AC\text{ subtends a right angle at the circumference,}
\displaystyle \therefore AC\text{ is the diameter of the circle.}
\\

\displaystyle \textbf{Question 58: }\text{In a cyclic quadrilateral }ABCD,\ \angle A:\angle C=3:1\text{ and }\angle B:\angle D=1:5.
\displaystyle \text{Find each angle of the quadrilateral.}
\displaystyle \text{Answer:}
\displaystyle \angle A:\angle C=3:1
\displaystyle \therefore \angle A=3x,\ \angle C=x
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle A+\angle C=180^\circ
\displaystyle 3x+x=180^\circ
\displaystyle 4x=180^\circ
\displaystyle \therefore x=45^\circ
\displaystyle \therefore \angle A=135^\circ,\ \angle C=45^\circ
\displaystyle \angle B:\angle D=1:5
\displaystyle \therefore \angle B=y,\ \angle D=5y
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle B+\angle D=180^\circ
\displaystyle y+5y=180^\circ
\displaystyle 6y=180^\circ
\displaystyle \therefore y=30^\circ
\displaystyle \therefore \angle B=30^\circ,\ \angle D=150^\circ
\\

\displaystyle \textbf{Question 59: }\text{The given figure shows a circle with centre }O\text{ and }\angle ABP=42^\circ.
\displaystyle \text{Find (i) }\angle PQB\text{ and (ii) }\angle QPB+\angle PBQ. c72\displaystyle \text{Answer:}
\displaystyle \text{Join }A\text{ and }P.
\displaystyle \text{(i) Since }AB\text{ is a diameter,}
\displaystyle \angle APB=90^\circ\text{ (angle in a semicircle)}c711
\displaystyle \text{In }\triangle APB,
\displaystyle \angle BAP=180^\circ-\angle APB-\angle ABP
\displaystyle \angle BAP=180^\circ-90^\circ-42^\circ=48^\circ
\displaystyle \angle PQB=\angle PAB\text{ (angles in the same segment subtended by chord }PB\text{)}
\displaystyle \therefore \angle PQB=48^\circ
\displaystyle \text{(ii) In }\triangle PBQ,
\displaystyle \angle QPB+\angle PBQ+\angle PQB=180^\circ
\displaystyle \angle QPB+\angle PBQ=180^\circ-48^\circ
\displaystyle \therefore \angle QPB+\angle PBQ=132^\circ
\\

\displaystyle \textbf{Question 60: }\text{In the given figure, }M\text{ is the centre of the circle. Chords }AB\text{ and }CD
\displaystyle \text{are perpendicular to each other. If }\angle MAD=x\text{ and }\angle BAC=y,\text{ find:}
\displaystyle \text{(i) }\angle AMD\text{ in terms of }x\quad\text{(ii) }\angle ABD\text{ in terms of }y
\displaystyle \text{(iii) Prove that }x=y. c71\displaystyle \text{Answer:}
\displaystyle \text{Let chords }AB\text{ and }CD\text{ intersect at }L.
\displaystyle \text{(i) }MA=MD\text{ (radii of the same circle)}
\displaystyle \therefore \angle MAD=\angle MDA=x\text{ (angles opposite equal sides)}
\displaystyle \text{In }\triangle AMD,
\displaystyle \angle AMD+\angle MAD+\angle MDA=180^\circ
\displaystyle \angle AMD+x+x=180^\circ
\displaystyle \therefore \angle AMD=180^\circ-2x
\displaystyle \text{Also, }\angle AMD=2\angle ABD
\displaystyle \text{(angle at the centre is twice the angle at the circumference subtended by chord }AD\text{)}
\displaystyle \angle ABD=\frac{1}{2}(180^\circ-2x)
\displaystyle \therefore \angle ABD=90^\circ-x\qquad\ldots\text{(i)}
\displaystyle \text{(ii) Since }AB\perp CD,\ \angle ALC=90^\circ.
\displaystyle \text{In }\triangle ALC,
\displaystyle \angle BAC+\angle ACD=90^\circ
\displaystyle y+\angle ACD=90^\circ
\displaystyle \therefore \angle ACD=90^\circ-y
\displaystyle \angle ABD=\angle ACD\text{ (angles in the same segment subtended by chord }AD\text{)}
\displaystyle \therefore \angle ABD=90^\circ-y\qquad\ldots\text{(ii)}
\displaystyle \text{(iii) From (i) and (ii),}
\displaystyle 90^\circ-x=90^\circ-y
\displaystyle \therefore x=y
\\

\displaystyle \textbf{Question 61: }\text{In a circle with centre }O,\text{ a cyclic quadrilateral }ABCD\text{ is drawn with }AB
\displaystyle \text{as a diameter and }CD\text{ equal to the radius of the circle. If }AD\text{ and }BC,\text{ when produced,}
\displaystyle \text{meet at }P,\text{ prove that }\angle APB=60^\circ. c710\displaystyle \text{Answer:}
\displaystyle OD=OC=CD\text{ (radii of the circle and }CD\text{ equals the radius)}
\displaystyle \therefore \triangle OCD\text{ is an equilateral triangle.}
\displaystyle \therefore \angle ODC=60^\circ
\displaystyle OA=OD\text{ (radii of the same circle)}
\displaystyle \therefore \angle OAD=\angle ODA\text{ (angles opposite equal sides)}
\displaystyle \angle ADC=\angle ADO+\angle ODC
\displaystyle \therefore \angle ADC=\angle OAD+60^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADC+\angle ABC=180^\circ
\displaystyle \angle OAD+60^\circ+\angle ABC=180^\circ
\displaystyle \therefore \angle OAD+\angle ABC=120^\circ
\displaystyle \text{Since }A,O,B\text{ are collinear and }A,D,P\text{ are collinear,}
\displaystyle \angle PAB=\angle OAD
\displaystyle \text{Since }B,C,P\text{ are collinear, }\angle ABP=\angle ABC.
\displaystyle \therefore \angle PAB+\angle ABP=120^\circ
\displaystyle \text{In }\triangle APB,
\displaystyle \angle APB=180^\circ-\angle PAB-\angle ABP
\displaystyle \angle APB=180^\circ-120^\circ
\displaystyle \therefore \angle APB=60^\circ
\\

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