\displaystyle \textbf{Question 1: }\text{Find the degree corresponding to the following radian measures.}
\displaystyle \text{Use }\pi=\frac{22}{7}.
\displaystyle \text{(i) }\left(\frac{9\pi}{5}\right)^c\qquad \text{(ii) }\left(\frac{-5\pi}{6}\right)^c\qquad \text{(iii) }\left(\frac{18\pi}{5}\right)^c
\displaystyle \text{(iv) }(-3)^c\qquad \text{(v) }11^c\qquad \text{(vi) }1^c
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\left(\frac{9\pi}{5}\right)^c=\left(\frac{9\pi}{5}\times\frac{180}{\pi}\right)^\circ=324^\circ
\displaystyle \text{(ii) }\left(\frac{-5\pi}{6}\right)^c=\left(\frac{-5\pi}{6}\times\frac{180}{\pi}\right)^\circ=-150^\circ
\displaystyle \text{(iii) }\left(\frac{18\pi}{5}\right)^c=\left(\frac{18\pi}{5}\times\frac{180}{\pi}\right)^\circ=648^\circ
\displaystyle \text{(iv) }(-3)^c=\left(-3\times\frac{180}{\pi}\right)^\circ=\left(-3\times\frac{180\times7}{22}\right)^\circ=-171\frac{18}{22}^\circ
\displaystyle =-171^\circ\left(\frac{18}{22}\times60\right)'=-171^\circ49'\left(\frac{2}{22}\times60\right)''=-171^\circ49'5''
\displaystyle \text{(v) }11^c=\left(11\times\frac{180\times7}{22}\right)^\circ=630^\circ
\displaystyle \text{(vi) }1^c=\left(1\times\frac{180\times7}{22}\right)^\circ=57\frac{6}{22}^\circ
\displaystyle =57^\circ\left(\frac{6}{22}\times60\right)'=57^\circ16'\left(\frac{8}{22}\times60\right)''=57^\circ16'22''
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the radian measure corresponding to the following degree measures:}
\displaystyle \text{(i) }300^\circ\qquad \text{(ii) }35^\circ\qquad \text{(iii) }-56^\circ\qquad \text{(iv) }135^\circ\qquad \text{(v) }-300^\circ
\displaystyle \text{(vi) }7^\circ30'\qquad \text{(vii) }125^\circ30'\qquad \text{(viii) }-47^\circ30'
\displaystyle \text{Answer:}
\displaystyle \text{(i) }300^\circ=\left(300\times\frac{\pi}{180}\right)^c=\left(\frac{5\pi}{3}\right)^c
\displaystyle \text{(ii) }35^\circ=\left(35\times\frac{\pi}{180}\right)^c=\left(\frac{7\pi}{36}\right)^c
\displaystyle \text{(iii) }-56^\circ=\left(-56\times\frac{\pi}{180}\right)^c=\left(\frac{-14\pi}{45}\right)^c
\displaystyle \text{(iv) }135^\circ=\left(135\times\frac{\pi}{180}\right)^c=\left(\frac{3\pi}{4}\right)^c
\displaystyle \text{(v) }-300^\circ=\left(-300\times\frac{\pi}{180}\right)^c=\left(\frac{-5\pi}{3}\right)^c
\displaystyle \text{(vi) }7^\circ30'=\left(7+\frac{30}{60}\right)^\circ=\frac{15}{2}^\circ
\displaystyle =\left(\frac{15}{2}\times\frac{\pi}{180}\right)^c=\left(\frac{\pi}{24}\right)^c
\displaystyle \text{(vii) }125^\circ30'=\left(125+\frac{30}{60}\right)^\circ=\frac{251}{2}^\circ
\displaystyle =\left(\frac{251}{2}\times\frac{\pi}{180}\right)^c=\left(\frac{251\pi}{360}\right)^c
\displaystyle \text{(viii) }-47^\circ30'=\left(-47-\frac{30}{60}\right)^\circ=\frac{-95}{2}^\circ
\displaystyle =\left(\frac{-95}{2}\times\frac{\pi}{180}\right)^c=\left(\frac{-19\pi}{72}\right)^c
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The difference between the two acute angles of a right-angled triangle is}
\displaystyle \frac{2\pi}{5}\text{ radians. Express the angles in degrees.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x_1\text{ and }x_2\text{ be the two acute angles of the right-angled triangle.}
\displaystyle \therefore x_1-x_2=\frac{2\pi}{5}\qquad\ldots\text{(i)}
\displaystyle \text{Also, }x_1+x_2=\frac{\pi}{2}\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle 2x_1=\left(\frac{2}{5}+\frac{1}{2}\right)\pi=\frac{9\pi}{10}
\displaystyle \Rightarrow x_1=\frac{9\pi}{20}=\frac{9}{20}\times180^\circ=81^\circ
\displaystyle \therefore x_2=90^\circ-81^\circ=9^\circ
\displaystyle \therefore \text{The angles are }81^\circ\text{ and }9^\circ.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the magnitude, in radians and degrees, of the interior angle of:}
\displaystyle \text{(i) Pentagon}\qquad\text{(ii) Octagon}\qquad\text{(iii) Heptagon}\qquad\text{(iv) Duodecagon}
\displaystyle \text{Answer:}
\displaystyle \text{Interior angle of a regular polygon with }n\text{ sides}=\left(\frac{2n-4}{n}\right)\times90^\circ
\displaystyle \text{(i) Pentagon}
\displaystyle \text{Interior angle}=\left(\frac{2\times5-4}{5}\right)\times90^\circ=108^\circ
\displaystyle \therefore \text{Interior angle in radians}=\left(108\times\frac{\pi}{180}\right)^c=\left(\frac{3\pi}{5}\right)^c
\displaystyle \text{(ii) Octagon}
\displaystyle \text{Interior angle}=\left(\frac{2\times8-4}{8}\right)\times90^\circ=135^\circ
\displaystyle \therefore \text{Interior angle in radians}=\left(135\times\frac{\pi}{180}\right)^c=\left(\frac{3\pi}{4}\right)^c
\displaystyle \text{(iii) Heptagon}
\displaystyle \text{Interior angle}=\left(\frac{2\times7-4}{7}\right)\times90^\circ=\frac{900}{7}^\circ=128^\circ34'17''
\displaystyle \therefore \text{Interior angle in radians}=\left(\frac{900}{7}\times\frac{\pi}{180}\right)^c=\left(\frac{5\pi}{7}\right)^c
\displaystyle \text{(iv) Duodecagon}
\displaystyle \text{Interior angle}=\left(\frac{2\times12-4}{12}\right)\times90^\circ=150^\circ
\displaystyle \therefore \text{Interior angle in radians}=\left(150\times\frac{\pi}{180}\right)^c=\left(\frac{5\pi}{6}\right)^c
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The angles of a quadrilateral are in A.P., and the greatest angle is }120^\circ.
\displaystyle \text{Express the angles in radians.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the angles of the quadrilateral, in degrees, be }a-3d,\ a-d,\ a+d\text{ and }a+3d.
\displaystyle \text{The sum of the angles of a quadrilateral is }360^\circ.
\displaystyle \therefore (a-3d)+(a-d)+(a+d)+(a+3d)=360^\circ
\displaystyle \Rightarrow 4a=360^\circ
\displaystyle \Rightarrow a=90^\circ
\displaystyle \text{Since the greatest angle is }120^\circ,
\displaystyle a+3d=120^\circ
\displaystyle \Rightarrow 90^\circ+3d=120^\circ
\displaystyle \Rightarrow 3d=30^\circ
\displaystyle \Rightarrow d=10^\circ
\displaystyle \therefore \text{The angles are }60^\circ,\ 80^\circ,\ 100^\circ\text{ and }120^\circ.
\displaystyle 60^\circ=60\times\frac{\pi}{180}=\frac{\pi}{3}\text{ radians}
\displaystyle 80^\circ=80\times\frac{\pi}{180}=\frac{4\pi}{9}\text{ radians}
\displaystyle 100^\circ=100\times\frac{\pi}{180}=\frac{5\pi}{9}\text{ radians}
\displaystyle 120^\circ=120\times\frac{\pi}{180}=\frac{2\pi}{3}\text{ radians}
\displaystyle \therefore \text{The angles in radians are }\frac{\pi}{3},\ \frac{4\pi}{9},\ \frac{5\pi}{9}\text{ and }\frac{2\pi}{3}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The angles of a triangle are in A.P., and the number of degrees in the least}
\displaystyle \text{angle is to the number of degrees in the mean angle as }1:120.\text{ Find the angles in radians.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three angles, in degrees, be }a-d,\ a\text{ and }a+d.
\displaystyle \text{Since the sum of the angles of a triangle is }180^\circ,
\displaystyle (a-d)+a+(a+d)=180^\circ
\displaystyle \Rightarrow 3a=180^\circ
\displaystyle \Rightarrow a=60^\circ
\displaystyle \text{Given that }\frac{\text{least angle}}{\text{mean angle}}=\frac{1}{120},
\displaystyle \frac{60-d}{60}=\frac{1}{120}
\displaystyle \Rightarrow 120(60-d)=60
\displaystyle \Rightarrow 7200-120d=60
\displaystyle \Rightarrow 120d=7140
\displaystyle \Rightarrow d=\frac{119}{2}^\circ=59^\circ30'
\displaystyle \therefore \text{The least angle}=60^\circ-59^\circ30'=0^\circ30'=\frac{1}{2}^\circ
\displaystyle \text{The mean angle}=60^\circ
\displaystyle \text{The greatest angle}=60^\circ+59^\circ30'=119^\circ30'=\frac{239}{2}^\circ
\displaystyle \frac{1}{2}^\circ=\frac{1}{2}\times\frac{\pi}{180}=\frac{\pi}{360}\text{ radians}
\displaystyle 60^\circ=60\times\frac{\pi}{180}=\frac{\pi}{3}\text{ radians}
\displaystyle \frac{239}{2}^\circ=\frac{239}{2}\times\frac{\pi}{180}=\frac{239\pi}{360}\text{ radians}
\displaystyle \therefore \text{The angles are }\frac{\pi}{360},\ \frac{\pi}{3}\text{ and }\frac{239\pi}{360}\text{ radians.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The interior angle of one regular polygon is to that of another regular polygon}
\displaystyle \text{as }3:2,\text{ and the number of sides of the first polygon is twice that of the second.}
\displaystyle \text{Determine the number of sides of the two polygons.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }n\text{ and }m\text{ be the numbers of sides of the first and second polygons respectively.}
\displaystyle \text{Interior angle of a regular polygon with }n\text{ sides}=\left(\frac{2n-4}{n}\right)\times90^\circ
\displaystyle \therefore \frac{\left(\frac{2n-4}{n}\right)\times90^\circ}{\left(\frac{2m-4}{m}\right)\times90^\circ}=\frac{3}{2}
\displaystyle \Rightarrow \frac{m}{n}\left(\frac{2n-4}{2m-4}\right)=\frac{3}{2}\qquad\ldots\text{(i)}
\displaystyle \text{Also, }n=2m.
\displaystyle \text{Substituting }n=2m\text{ in (i),}
\displaystyle \frac{m}{2m}\left(\frac{4m-4}{2m-4}\right)=\frac{3}{2}
\displaystyle \Rightarrow \frac{4m-4}{2m-4}=3
\displaystyle \Rightarrow 4m-4=6m-12
\displaystyle \Rightarrow 2m=8
\displaystyle \Rightarrow m=4
\displaystyle \therefore n=2m=2\times4=8
\displaystyle \therefore \text{The first polygon has }8\text{ sides and the second polygon has }4\text{ sides.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The angles of a triangle are in A.P. such that the greatest angle is five times}
\displaystyle \text{the least angle. Find the angles in radians.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three angles, in degrees, be }A=a-d,\ B=a\text{ and }C=a+d.
\displaystyle \text{Since the sum of the angles of a triangle is }180^\circ,
\displaystyle (a-d)+a+(a+d)=180^\circ
\displaystyle \Rightarrow 3a=180^\circ
\displaystyle \Rightarrow a=60^\circ
\displaystyle \text{Given that the greatest angle is five times the least angle,}
\displaystyle C=5A
\displaystyle \Rightarrow a+d=5(a-d)
\displaystyle \Rightarrow a+d=5a-5d
\displaystyle \Rightarrow 6d=4a
\displaystyle \Rightarrow d=\frac{2a}{3}=\frac{2}{3}\times60^\circ=40^\circ
\displaystyle \therefore A=a-d=60^\circ-40^\circ=20^\circ
\displaystyle B=a=60^\circ
\displaystyle C=a+d=60^\circ+40^\circ=100^\circ
\displaystyle 20^\circ=20\times\frac{\pi}{180}=\frac{\pi}{9}\text{ radians}
\displaystyle 60^\circ=60\times\frac{\pi}{180}=\frac{\pi}{3}\text{ radians}
\displaystyle 100^\circ=100\times\frac{\pi}{180}=\frac{5\pi}{9}\text{ radians}
\displaystyle \therefore \text{The angles are }\frac{\pi}{9},\ \frac{\pi}{3}\text{ and }\frac{5\pi}{9}\text{ radians.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The numbers of sides of two regular polygons are in the ratio }
\displaystyle 5:4,\text{ and the}  \ \text{difference between their interior angles is }9^\circ. \\ \text{ Find the numbers of sides of the polygons.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }n\text{ and }m\text{ be the numbers of sides of the two regular polygons respectively.}
\displaystyle \therefore \frac{n}{m}=\frac{5}{4}
\displaystyle \Rightarrow n=\frac{5m}{4}\qquad\ldots\text{(i)}
\displaystyle \text{Interior angle of a regular polygon with }n\text{ sides}=\left(\frac{2n-4}{n}\right)90^\circ
\displaystyle \text{Since the difference between the interior angles is }9^\circ,
\displaystyle \left(\frac{2n-4}{n}\right)90^\circ-\left(\frac{2m-4}{m}\right)90^\circ=9^\circ
\displaystyle \Rightarrow \frac{2n-4}{n}-\frac{2m-4}{m}=\frac{1}{10}
\displaystyle \Rightarrow \frac{m(2n-4)-n(2m-4)}{mn}=\frac{1}{10}
\displaystyle \Rightarrow \frac{2mn-4m-2mn+4n}{mn}=\frac{1}{10}
\displaystyle \Rightarrow \frac{4(n-m)}{mn}=\frac{1}{10}
\displaystyle \text{Substituting }n=\frac{5m}{4}\text{ from (i),}
\displaystyle \frac{4\left(\frac{5m}{4}-m\right)}{\frac{5m}{4}\times m}=\frac{1}{10}
\displaystyle \Rightarrow \frac{4\left(\frac{m}{4}\right)}{\frac{5m^2}{4}}=\frac{1}{10}
\displaystyle \Rightarrow \frac{4}{5m}=\frac{1}{10}
\displaystyle \Rightarrow 5m=40
\displaystyle \Rightarrow m=8
\displaystyle \therefore n=\frac{5}{4}\times8=10
\displaystyle \therefore \text{The two polygons have }10\text{ sides and }8\text{ sides respectively.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A railway curve is to be laid out along a circle. What radius should be used if}
\displaystyle \text{the track is to change direction by }25^\circ\text{ over a distance of }40\text{ metres?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ metres be the radius of the circular curve.}
\displaystyle \text{Arc length }s=40\text{ m}
\displaystyle \theta=25^\circ=25\times\frac{\pi}{180}=\frac{5\pi}{36}\text{ radians}
\displaystyle \text{Using }s=r\theta,
\displaystyle 40=r\times\frac{5\pi}{36}
\displaystyle \Rightarrow r=\frac{40\times36}{5\pi}
\displaystyle \Rightarrow r=\frac{288}{\pi}\text{ m}
\displaystyle \Rightarrow r=91.67\text{ m (approx.)}
\displaystyle \therefore \text{The required radius is }\frac{288}{\pi}\text{ m, or }91.67\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the length which, at a distance of }5280\text{ m, subtends an angle of }1'
\displaystyle \text{at the eye.}
\displaystyle \text{Answer:}
\displaystyle \text{Given radius }r=5280\text{ m}
\displaystyle \angle AOB=1'=\frac{1}{60}^\circ=\frac{1}{60}\times\frac{\pi}{180}=\frac{\pi}{10800}\text{ radians}
\displaystyle \text{Using }\theta=\frac{\text{arc length}}{\text{radius}},
\displaystyle \frac{\pi}{10800}=\frac{l}{5280}
\displaystyle \Rightarrow l=\frac{5280\pi}{10800}=\frac{22\pi}{45}\text{ m}
\displaystyle \text{Using }\pi=\frac{22}{7},
\displaystyle l=\frac{22}{45}\times\frac{22}{7}=1.5365\text{ m (approx.).}
\displaystyle \therefore \text{The required length is }1.5365\text{ m (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A wheel makes }360\text{ revolutions in a minute. Through how many radians}
\displaystyle \text{does it turn in }1\text{ second?}
\displaystyle \text{Answer:}
\displaystyle \text{Given the wheel makes }360\text{ revolutions per minute.}
\displaystyle \therefore \text{Revolutions per second}=\frac{360}{60}=6
\displaystyle \text{In one revolution, the wheel turns through }360^\circ.
\displaystyle \therefore \text{In one second, it turns through }6\times360^\circ=2160^\circ.
\displaystyle \text{In radians, }2160^\circ=2160\times\frac{\pi}{180}=12\pi\text{ radians.}
\displaystyle \therefore \text{The wheel turns through }12\pi\text{ radians in }1\text{ second.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the angle, in radians, through which a pendulum swings if its length is }75\text{ cm}
\displaystyle \text{and the tip describes an arc of length (i) }10\text{ cm}\text{ (ii) }15\text{ cm}\text{ (iii) }21\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle \text{Length of the pendulum (radius) }=75\text{ cm}
\displaystyle \text{(i) Arc length}=10\text{ cm}
\displaystyle \therefore \theta=\frac{\text{arc length}}{\text{radius}}=\frac{10}{75}=\frac{2}{15}\text{ radians}
\displaystyle \text{(ii) Arc length}=15\text{ cm}
\displaystyle \therefore \theta=\frac{\text{arc length}}{\text{radius}}=\frac{15}{75}=\frac{1}{5}\text{ radians}
\displaystyle \text{(iii) Arc length}=21\text{ cm}
\displaystyle \therefore \theta=\frac{\text{arc length}}{\text{radius}}=\frac{21}{75}=\frac{7}{25}\text{ radians}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The radius of a circle is }30\text{ cm. Find the length of an arc of the circle if}
\displaystyle \text{the length of its chord is }30\text{ cm.}
\displaystyle \text{Answer:}
\displaystyle OA=OB=\text{radius}=30\text{ cm}=0.3\text{ m}
\displaystyle AB=\text{chord}=30\text{ cm}=0.3\text{ m}
\displaystyle \therefore \triangle AOB\text{ is equilateral.}
\displaystyle \therefore \angle AOB=60^\circ=\frac{\pi}{3}\text{ radians}
\displaystyle \text{Let the arc length }AB=l.
\displaystyle \theta=\frac{l}{r}
\displaystyle \Rightarrow \frac{\pi}{3}=\frac{l}{0.3}
\displaystyle \Rightarrow l=0.3\times\frac{\pi}{3}=0.1\pi\text{ m}
\displaystyle =10\pi\text{ cm}
\displaystyle \therefore \text{The required arc length is }10\pi\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A railway train is travelling on a circular curve of radius }1500\text{ m at a}
\displaystyle \text{speed of }66\text{ km/h. Through what angle does it turn in }10\text{ seconds?}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circular curve }r=1500\text{ m}
\displaystyle \text{Speed of the train}=66\text{ km/h}
\displaystyle =66\times\frac{1000}{3600}\text{ m/s}=\frac{55}{3}\text{ m/s}
\displaystyle \text{Distance travelled in }10\text{ seconds}=\frac{55}{3}\times10=\frac{550}{3}\text{ m}
\displaystyle \therefore \text{Arc length }s=\frac{550}{3}\text{ m}
\displaystyle \text{Using }\theta=\frac{s}{r},
\displaystyle \theta=\frac{\frac{550}{3}}{1500}=\frac{550}{4500}=\frac{11}{90}\text{ radians}
\displaystyle \therefore \text{The train turns through }\frac{11}{90}\text{ radians in }10\text{ seconds.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the distance from the eye at which a coin of diameter }2\text{ cm should be}
\displaystyle \text{held so as to conceal the full moon whose angular diameter is }31'.
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ metres be the distance of the coin from the eye.}
\displaystyle \theta=31'=\frac{31}{60}^\circ
\displaystyle =\frac{31}{60}\times\frac{\pi}{180}=\frac{31\pi}{10800}\text{ radians}
\displaystyle \text{Diameter of the coin}=2\text{ cm}=0.02\text{ m}
\displaystyle \text{Using }\theta=\frac{s}{r},
\displaystyle \frac{31\pi}{10800}=\frac{0.02}{r}
\displaystyle \Rightarrow r=\frac{0.02\times10800}{31\pi}
\displaystyle \Rightarrow r=\frac{216}{31\pi}\text{ m}
\displaystyle \Rightarrow r=2.22\text{ m (approx.)}
\displaystyle \therefore \text{The coin should be held approximately }2.22\text{ m from the eye.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the diameter of the sun in km, supposing that it subtends an angle of }32'
\displaystyle \text{at the eye of an observer. Given that the distance of the sun is }91\times10^6\text{ km.}
\displaystyle \text{Answer:}
\displaystyle \theta=32'=\frac{32}{60}^\circ=\frac{32}{60}\times\frac{\pi}{180}=\frac{2\pi}{675}\text{ radians}
\displaystyle \text{Distance of the sun}=91\times10^6\text{ km}
\displaystyle \text{Using }\theta=\frac{\text{arc length}}{\text{radius}},
\displaystyle \frac{2\pi}{675}=\frac{AB}{91\times10^6}
\displaystyle \Rightarrow AB=\frac{2\pi}{675}\times91\times10^6
\displaystyle \Rightarrow AB=8.47\times10^5\text{ km (approx.)}
\displaystyle \therefore \text{The diameter of the sun is approximately }847407.4\text{ km.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If the arcs of the same length in two circles subtend angles }65^\circ\text{ and }110^\circ
\displaystyle \text{at the centres, find the ratio of their radii.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }C_1\text{ and }C_2\text{ be the two circles having the same arc length }l.
\displaystyle \text{Let their radii be }r\text{ and }R\text{ respectively.}
\displaystyle \theta_1=65^\circ=\frac{65\pi}{180}\text{ radians}
\displaystyle \theta_2=110^\circ=\frac{110\pi}{180}\text{ radians}
\displaystyle \text{Using }\theta=\frac{\text{arc length}}{\text{radius}},
\displaystyle \theta_1=\frac{l}{r}\Rightarrow r=\frac{l}{\theta_1}
\displaystyle \theta_2=\frac{l}{R}\Rightarrow R=\frac{l}{\theta_2}
\displaystyle \therefore \frac{r}{R}=\frac{\frac{l}{\theta_1}}{\frac{l}{\theta_2}}=\frac{\theta_2}{\theta_1}
\displaystyle =\frac{\frac{110\pi}{180}}{\frac{65\pi}{180}}=\frac{22}{13}
\displaystyle \therefore r:R=22:13
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find the degree measure of the angle subtended at the centre of a circle of radius}
\displaystyle 100\text{ cm by an arc of length }22\text{ cm. Use }\pi=\frac{22}{7}.
\displaystyle \text{Answer:}
\displaystyle \text{Arc }AB=22\text{ cm}
\displaystyle OA=OB=r=100\text{ cm}
\displaystyle \text{Let }\theta\text{ be the angle subtended by arc }AB\text{ at the centre }O.
\displaystyle \text{Using }\theta=\frac{\text{arc length}}{\text{radius}},
\displaystyle \theta=\frac{22}{100}=\frac{11}{50}\text{ radians}
\displaystyle \therefore \theta=\frac{11}{50}\times\frac{180}{\pi}^\circ
\displaystyle =\frac{11}{50}\times\frac{180\times7}{22}^\circ=12.6^\circ
\displaystyle =12^\circ+0.6^\circ=12^\circ36'
\displaystyle \therefore \text{The required angle is }12^\circ36'.
\displaystyle \\

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