Question 41: In the given figure, $AB$ is the diameter of the circle with center $O$. $DO \parallel CB$ and $\angle DCB = 120^o$. Calculate (i) $\angle DAB$ (ii) $\angle DBA$ (iii) $\angle DBC$(iv) $\angle ADC$. Show that $\triangle AOD$ is an equilateral triangle.

(i) $ABCD$ is a cyclic quadrilateral

$\therefore \angle DAB + \angle DCB = 180^o$

$\Rightarrow \angle DAB = 60^o$

(ii) $\angle ADB = 90^o$ (angle in the  semicircle)

In $\triangle ADO$,

$\angle ADO = \angle AOD = 60^o$

$\therefore \angle ODB = 30^o$

In $\triangle \angle ODB, OD = OB$

$\Rightarrow \angle DBA = 30^o$

(iii) $\angle ODB = 30^o$

$\therefore \angle DBC = 30^o$ (alternate angles)

(iv) $\angle ABC = 60^o$

$\angle ADC + \angle ABC = 180^o$ (cyclic quadrilateral)

$\angle ADC = 180^o-60^o = 120^o$

In $\triangle AOD$,

$OD = OA$ (Radius of the same circle)

$\therefore \angle ADO = \angle AOD = 60^o$

Therefore all angles are $60^o$. Hence $\triangle AOD$ is an equilateral triangle.

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Question 42: In the given figure $I$ is the incenter of the $\triangle ABC$. $BI$ when produced meets the circumcenter of the the $\triangle ABC$ at $D$. Given $\angle BAC = 55^o$ and $\angle ACB = 65^o$. Calculate: (i) $\angle DCA$ (ii) $\angle DAC$ (iii) $\angle DCI$ (iv) $\angle AIC$

(i) $IB$ is bisector of  $\angle ABC$

$\angle ABD = \frac{1}{2} \angle ABC$

$\angle ABC = 180^o-55^o-65^o=60^o$

$\therefore \angle ABD = 30^o$

(ii) $\angle DAC = \angle CBD = 30^o$ (angles in the same segment)

(iii) $\angle ACI = \frac{1}{2} \angle ACB$

$CI \ bisects \ \angle ACB$

$\therefore \angle ACI = \frac{65}{2} = 32.5^o$

(iv) $\angle IAC = \frac{1}{2} \angle BAC = \frac{55}{2} = 27.5^o$

$AI$ bisects $\angle BAC$

$\therefore \angle AIC = 180^o - \angle IAC- \angle ICA = 180^o-27.5^o-32.5^o = 120^o$

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Question 43: A $\triangle ABC$ is inscribed in a circle. The bisector of $\angle BAC, \angle ABC$ and $\angle ACB$ meet the circumference of the triangle at points $P, Q$ and $R$ respectively. Prove that (i) $\angle ABC = 2 \angle APQ$ (ii) $\angle ACB = 2 \angle APR$ (iii) $\angle QPR = 90^o- \frac{1}{2} \angle BAC$

(i) $BQ$ bisects $\angle ABC$

$\therefore \angle ABQ = \frac{1}{2} \angle ABC$

$\angle APQ = \angle ABQ$ (angle in same segment)

$\therefore \angle ABC = 2 \angle ABQ = 2 \angle APQ$ … … … … (i)

(ii) $CR$ bisects $\angle ACB$

$\therefore \angle ACR = \frac{1}{2} \angle ABC$

and $\angle ACR = \angle APR$ (angles in the same segment)

$\therefore \angle ACB = 2 \angle APR$ … … … … (ii)

(iii) Adding (i) and (ii)  we get

$\angle ABC + \angle ACB = 2 (\angle APQ + \angle APR) = 2 \angle QPR$

$\Rightarrow 180^o - \angle BAC = 2 \angle QPR$

$\Rightarrow \angle QPR = 90^o-\frac{1}{2} \angle BAC$

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Question 44: Calculate angles $x, y, z$ if: $\frac{x}{3}= \frac{y}{4}=\frac{z}{5}$

$\frac{x}{3}= \frac{y}{4}=\frac{z}{5} = k \ (say)$

$\Rightarrow x = 3k, y = 4k \ and \ z = 5k$

$\angle ABC + \angle ADC = 180^o$

$\angle PBC = 180^o-3k-4k = 180^o-7k$

$\therefore ABC = 7k$

Similarly $\angle ADC = 180^o -(180^o-3k-5k) = 8k$

$\therefore \angle ABC + \angle ADC = 7k+8k = 15k$

$15 = 180^o \Rightarrow k = 12$

Therefore $x = 36^o, y = 48^o \ and \ z = 60^o$

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Question 45: In the given figure $AB = AC = CD$ and $\angle ADC = 38^o$, calculate (i) $\angle ABC$ (ii) $\angle BEC$ [1995]

(i) $AC = CD$

$\angle CAD = \angle CDA = 38^o$

$\angle ACD = 180^o-38^o - 38^o = 104^o$

$\angle ACB = 180^o-104^o = 76^o$

$AB = AC$

$\angle ABC = \angle ACB = 76^o$

(ii) $\angle BAC = 180^o-76^o-76^o = 38^o$

$\angle BAC = \angle BEC = 38^o$ (angles in the same segment)

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Question 46: In the given figure $AC$ is the diameter of the circle with center $O$. Chord $BD \perp AC$. Write down the angles $p, q$ and $r$ in terms of $x$. [1996]

$\angle AOB = 2 \angle ACB = 2 \angle ADB$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$x = 2q \Rightarrow q=\frac{x}{2}$

$\angle ADB = \frac{x}{2}$

$\angle ADC = 90^o$ (angles in a semicircle)

$r +\frac{x}{2} = 90^o \Rightarrow r = 90-\frac{x}{2}$

Now $\angle DAC = \angle DBC$ (angles in the same segment)

$p = 90^o-q \Rightarrow p = 90^o-\frac{x}{2}$

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Question 47:  In the given figure $AC$ is the diameter of the circle with center $O$. $CD \parallel BE &s=0$. $\angle AOB = 80^o$ and $\angle ACE = 110^o$. Calculate (i) $\angle BEC$ (ii) $\angle BCD$ (iii) $\angle CED$ [1998]

(i) $\angle BOC = 180^o-80^o=100^o$ (Since $AC$ is a straight line)

$\angle BOC = 2 \angle BEC$  (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\Rightarrow \angle BEC = \frac{100}{2} = 50^o$

(ii) $DC \parallel EB$

$\angle DCE = \angle BEC = 50^o$

$\angle AOB = 80^o$ (given)

$\Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\angle BCD = \angle ACB + \angle ACE + \angle DCE = 40^o+10^o+50^o = 100^o$

(iii) $\angle BED = 180^o-\angle BCD = 180^o-100^o = 80^o$ (cyclic quadrilateral)

$\angle CED + 50^o = 80^o \Rightarrow \angle CED = 30^o$

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Question 48: In the given figure, $AE$ is the diameter of the circle. Write down the numerical value of $\angle ABC + \angle CDE$. Give reasons for your answer. [1998]

$\angle AOC = \frac{180}{2} = 90^o$

$\angle AOC = 2 \angle AEC$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\Rightarrow \angle AEC = \frac{90}{2} = 45^o$

$ABCE$ is a cyclic quadrilateral

$\therefore \angle ABC + \angle AEC = 180 ^o$

$\Rightarrow \angle ABC = 180-45 = 135^o$

Similarly, $\angle CDE = 135^o$

$\therefore \angle ABC + \angle CDE = 135^o + 135^o = 270^o$

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Question 49: In the given figure $AOC$ is the diameter and $AC \parallel ED$. If $\angle CBE = 64^o$, calculate $\angle DEC$. [1991]

$\angle ABC = 90^o$ (angle in a semi circle)

$\angle ABE = 90^o-64^o = 26^o$

$\angle ABE = \angle ACE = 26^o$ (angles in the same segment)

$AC \parallel ED$

$\therefore \angle DEC = \angle ACE = 26^o$ (alternate angles)

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Question 50: Use the given figure below to find (i) $\angle BAD$ (ii) $\angle DQB$ [1987]

(i) In $\triangle ADP$

$\angle BAD = 180^o - 85^o - 40^o = 55^o$

(ii) $\angle ABC = 180^o - \angle ADC = 180^o - 85^o = 95^o$

$\angle AQB = 180^o - 95^o - 55^o = 30^o$

$\Rightarrow \angle DQB = 30^o$

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