Class 11: Measurement of Angles – Exercise 4.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the degree corresponding to the following radian measures $\displaystyle \text{Use } \pi = \frac{22}{7}$

$\displaystyle \text{(i) } \Big( \frac{9 \pi}{5} \Big)^c \hspace{1.0cm} \text{(ii) } \Big( \frac{-5 \pi}{6} \Big)^c \hspace{1.0cm} \text{(iii) } \Big( \frac{18 \pi}{5} \Big)^c \hspace{1.0cm} \text{(iv) } (-3)^c \hspace{1.0cm} \text{(v) } 11^c \hspace{1.0cm} \text{(vi) } 1^c$

Answer:

$\displaystyle \text{(i) } \Big( \frac{9\pi}{5} \Big)^c = \Big( \frac{9\pi}{5} \times \frac{180}{\pi} \Big)^{\circ} = 324^{\circ}$

$\displaystyle \text{(ii) } \Big( \frac{-5\pi}{6} \Big)^c = \Big( \frac{-5\pi}{6} \times \frac{180}{\pi} \Big)^{\circ} = -150^{\circ}$

$\displaystyle \text{(iii) } \Big( \frac{18\pi}{5} \Big)^c = \Big( \frac{18\pi}{5} \times \frac{180}{\pi} \Big)^{\circ} = 648^{\circ}$

$\displaystyle \text{(iv) } (-3)^c = \Big( -3 \times \frac{180}{\pi} \Big)^{\circ} = \Big( -3 \times \frac{180}{22} \times 7 \Big)^{\circ} = \Big( -171 \frac{18}{22} \Big)^{\circ}$

$\displaystyle = -171^{\circ} \Big( \frac{18}{22} \times 60 \Big)' = -171^{\circ} 49' \Big( \frac{2}{22} \times 60 \Big)'' = - 171^{\circ} 49' 5''$

$\displaystyle \text{(v) } (11)^c =\Big( 11 \times \frac{180}{22} \times 7 \Big)^{\circ} = 630^{\circ}$

$\displaystyle \text{(vi) } (1)^c = \Big( 1 \times \frac{180}{22} \times 7 \Big) = 57^{\circ} \Big( \frac{6}{22} \times 60 \Big)' = 57^{\circ} 16' \Big( \frac{8}{22} \times 60 \Big)'' = 57^{\circ} 16' 21''$

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Question 2: Find the radian measure corresponding to the following degree measures:

$\displaystyle \text{(i) } 300^{\circ} \hspace{1.0cm} \text{(ii) } 35^{\circ} \hspace{1.0cm} \text{(iii) } -56^{\circ} \hspace{1.0cm} \text{(iv) } 135^{\circ} \hspace{1.0cm} \text{(v) } -300^{\circ} \hspace{1.0cm} \\ \\ \text{(vi) } 7^{\circ} 30' \hspace{1.0cm} \text{(vii) } 125^{\circ} 30' \hspace{1.0cm} \text{(viii) } -47^{\circ} 30'$

Answer:

$\displaystyle \text{(i) } 300^{\circ} = \Big( 300 \times \frac{\pi}{180} \Big)^c = \Big( \frac{5\pi}{3} \Big)^c$

$\displaystyle \text{(ii) } 35^{\circ} = \Big( 35 \times \frac{\pi}{180} \Big)^c = \Big( \frac{7\pi}{36} \Big)^c$

$\displaystyle \text{(iii) } -56^{\circ} = \Big( -56 \times \frac{\pi}{180} \Big)^c = \Big( \frac{-14\pi}{45} \Big)^c$

$\displaystyle \text{(iv) } 135^{\circ} = \Big( 135 \times \frac{\pi}{180} \Big)^c = \Big( \frac{3\pi}{4} \Big)^c$

$\displaystyle \text{(v) } -300^{\circ} = \Big( -300 \times \frac{\pi}{180} \Big)^c = \Big( \frac{-5\pi}{3} \Big)^c$

$\displaystyle \text{(vi) } 7^{\circ}30' = \Big( 7 \frac{30}{60} \Big)^{\circ} = \Big(7 \frac{1}{2} \Big)^{\circ} = \Big( \frac{15}{2} \times \frac{\pi}{180} \Big)^c = \Big( \frac{\pi}{24} \Big)^c$

$\displaystyle \text{(vii) } 125^{\circ}30' = \Big( 125 \frac{30}{60} \Big)^{\circ} = \Big(125 \frac{1}{2} \Big)^{\circ} = \Big( \frac{251}{2} \times \frac{\pi}{180} \Big)^c = \Big( \frac{251\pi}{360} \Big)^c$

$\displaystyle \text{(viii) } -47^{\circ}30' = \Big( -47 \frac{30}{60} \Big)^{\circ} = \Big(-47 \frac{1}{2} \Big)^{\circ} = \Big( \frac{-95}{2} \times \frac{\pi}{180} \Big)^c = \Big( \frac{19\pi}{72} \Big)^c$

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Question 3: The difference between the two acute angles of a right-angled triangle is $\displaystyle \frac{2 \pi}{5}$ radians. Express the angles in degrees.

Answer:

$\displaystyle \text{Let } x_1 \text{ and } x_2$ be two acute angles of a right angles triangle.

$\displaystyle \therefore x_1 - x_2 = \frac{2\pi}{5}$ … … … … … i)

Also in a right angled triangle

$\displaystyle x_1 + x_ 2 = \frac{\pi}{2}$ … … … … … ii)

Adding i) and ii) we get

$\displaystyle 2x_1 = \Big( \frac{2}{5} + \frac{1}{2} \Big) \pi$

$\displaystyle \Rightarrow 2 x_1 = \frac{9}{10} \pi$

$\displaystyle \Rightarrow x_1 = \frac{9}{20} \pi = \frac{9}{20} \times 180 = 81^{\circ}$

$\displaystyle \therefore x_2 = 90- 81 = 9^{\circ}$

$\displaystyle \text{Hence } x_1 = 81^{\circ} \text{ and } x_2 = 9^{\circ}$

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Question 4: One angle of a triangle is $\displaystyle \frac{2x}{3}$ grades and another is $\displaystyle \frac{3x}{2}$ degrees while the third angle is $\displaystyle \frac{\pi }{75}$ radians. Express all the angles in degrees.

Answer:

Let the three angles be $\displaystyle x_1, x_2 \text{ and } x_3$

$\displaystyle \therefore x_1 = \frac{2}{3} x \text{ gradiant } = \Big( \frac{2}{3} x \times \frac{90}{100} \Big)^{\circ} = \Big( \frac{3}{5} x \Big)^{\circ}$

$\displaystyle x_2 = \Big( \frac{3}{2} x \Big)^{\circ}$

$\displaystyle x_3 = \Big( \frac{\pi}{75} x \Big)^c = \Big( \frac{\pi}{75} x \times \frac{180}{\pi} \Big)^{\circ} = \Big( \frac{12}{5} x \Big)^{\circ}$

In the triangle

$\displaystyle x_1 + x_2 + x_3 = \pi$

$\displaystyle \Rightarrow \frac{3}{5} x + \frac{3}{2} x + \frac{12}{5} x = \pi$

$\displaystyle \Rightarrow \frac{6x+15x+24x}{10} = \pi$

$\displaystyle \Rightarrow \frac{45x}{10} = \pi \Rightarrow x = \frac{2}{9} \pi = 40^{\circ}$

$\displaystyle \therefore x_1 = \frac{3}{5} \times 40 = 24^{\circ} x_2 = \frac{3}{2} \times 40 = 60^{\circ} x_3 = \frac{12}{5} \times 40 = 96^{\circ}$

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Question 5: Find the magnitude, in radians and degrees, of interior angle of: i) pentagon ii) octagon (iii) heptagon iv) duodecagon

Answer:

Interior angles of a polygon with $\displaystyle n$ sides $\displaystyle = \Big( \frac{2n-4}{n} \Big) \times 90^{\circ}$

i) Pentagon

$\displaystyle \text{Interior angle } = \Big( \frac{2 \times 5-4}{5} \Big) \times 90^{\circ} = 108^{\circ}$

$\displaystyle \text{Therefore in radians, interior angle } = \Big( 108 \times \frac{\pi}{180} \Big) = \Big( \frac{3\pi}{5} \Big)^c$

ii) Octagon

$\displaystyle \text{Interior angle } = \Big( \frac{2 \times 8-4}{8} \Big) \times 90^{\circ} = 135^{\circ}$

$\displaystyle \text{Therefore in radians, interior angle } = \Big( 135 \times \frac{\pi}{180} \Big)^c = \Big( \frac{3\pi}{4} \Big)^c$

(iii) Heptagon

$\displaystyle \text{Interior angle } = \Big( \frac{2 \times 7-4}{7} \Big) \times 90^{\circ} = 128^{\circ} 34' 17''$

$\displaystyle \text{Therefore in radians, interior angle } = \Big( \frac{900}{7} \times \frac{\pi}{180} \Big)^c = \Big( \frac{5\pi}{7} \Big)^c$

iv) Duodecagon

$\displaystyle \text{Interior angle } = \Big( \frac{2 \times 12-4}{12} \Big) \times 90^{\circ} = 150^{\circ}$

$\displaystyle \text{Therefore in radians, interior angle } = \Big( 150 \times \frac{\pi}{180} \Big)^c = \Big( \frac{5\pi}{6} \Big)^c$

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Question 6: The angles of a quadrilateral are in A.P. and the greatest angles is $\displaystyle 120^{\circ}$ . Express the angles in radians.

Answer:

Let the angles of quadrilateral in degrees be $\displaystyle a-3d, a-d, a+d, a+3d$

We know, sum of angles in quadrilateral $\displaystyle = 360^{\circ}$

$\displaystyle \therefore (a-3d) + (a-d) + ( a+d) + ( a+3d) = 360$

$\displaystyle \Rightarrow 4a = 360 \Rightarrow 4 = 90^{\circ}$

Largest angle $\displaystyle a+3d = 120$

$\displaystyle \Rightarrow 90+ 3d = 120 \Rightarrow d = 10$

$\displaystyle \therefore$ angles are $\displaystyle 60^{\circ}, 80^{\circ}, 100^{\circ} \text{ and } 120^{\circ}$

$\displaystyle \text{In radians angles are } 60 \times \frac{\pi}{180} , 80 \times \frac{\pi}{180} , 100 \times \frac{\pi}{180}$ , $\displaystyle \text{and } 120 \times \frac{\pi}{180}$

$\displaystyle \Rightarrow \frac{\pi}{3} , \frac{4\pi}{9} , \frac{5\pi}{9}$ , $\displaystyle \text{and } \frac{2\pi}{3}$

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Question 7: The angles of a quadrilateral are in A.P. and the number of degrees in the least angles is to the number of degrees in the mean angle as $\displaystyle 1:120$ . Find the angles in radians.

Answer:

$\displaystyle \text{Let } A, B \text{ and } C$ be the three angles

It is given that $\displaystyle A, B \text{ and } C$ are in an AP

$\displaystyle \text{Let } A = a-d, \ B = a$ , $\displaystyle \text{and } C = a+d$

$\displaystyle \therefore A + B + C = 180^{\circ} \Rightarrow 3a = 180^{\circ} \Rightarrow a = 60^{\circ} \Rightarrow a \Big( \frac{\pi}{3} \Big)^c$

$\displaystyle \text{Given that } \frac{\text{least angle}}{\text{mean angle}} = \frac{1}{120}$

$\displaystyle \Rightarrow \frac{60-d}{60} = \frac{1}{120}$

$\displaystyle \Rightarrow 120 - 120d = 1$

$\displaystyle \Rightarrow d = \Big( \frac{119}{120} \Big)^{\circ} = \Big( \frac{119}{120} \times 60 \Big)' = \Big( \frac{119}{2} \Big)'$

$\displaystyle \therefore \ d \text{ in radians } = \Big( \frac{119}{2} \times \frac{\pi}{180} \Big)^c = \Big( \frac{119}{360} \Big)^c$

$\displaystyle \therefore A = \frac{\pi}{3} - \frac{119\pi}{360} = \Big( \frac{\pi}{360} \Big)^c$

$\displaystyle \text{and } C = \frac{\pi}{3} + \frac{119\pi}{360} = \Big( \frac{239\pi}{360} \Big)^c$

$\displaystyle \text{and } B = \Big( \frac{\pi}{3} \Big)^c$

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Question 8: The angle in a regular polygon is to that in another is $\displaystyle 3:2$ and the number of sides in first is twice that in second. Determine the number of sides of two polygons.

Answer:

$\displaystyle \text{Let } n \text{ and } m$ be the number of sides of two regular polygons respectively.

$\displaystyle \therefore \frac{\frac{2n-4}{n} \times 90^{\circ}}{\frac{2m-4}{m} \times 90^{\circ}} = \frac{3}{2}$

$\displaystyle \Rightarrow \frac{m}{n} \Big( \frac{2n-4}{2m-4} \Big) = \frac{3}{2}$ … … … … … i)

Also $\displaystyle \text{Given } n = 2m$

Substituting in i) we get

$\displaystyle \frac{m}{2m} \Big( \frac{4m-4}{2m-4} \Big) = \frac{3}{2}$

$\displaystyle \Rightarrow \frac{4m-4}{2m-4} = 3$

$\displaystyle \Rightarrow 4m - 4 = 6m - 12$

$\displaystyle \Rightarrow 2m = 8 \Rightarrow m = 4$

$\displaystyle \therefore n = 2 \times 4 = 8$

Hence $\displaystyle n = 8 \text{ and } m = 4$

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Question 9: The angles of a triangle are in A.P. such that the greatest angle is five times the least. Find the angles in radians.

Answer:

$\displaystyle \text{Let } A, B \text{ and } C$ be the three angles

$\displaystyle \text{Let } A = a - d, B = a \text{ and } C = a+d$

$\displaystyle \therefore a-d + a + a + d = 180^{\circ}$

$\displaystyle \Rightarrow 3a = 180^{\circ} \Rightarrow a = 60^{\circ}$

$\displaystyle \text{Given } C = 5 A$

$\displaystyle a+ d = 5 (a - d) \Rightarrow 6d = 4a$

$\displaystyle \therefore d = \frac{4}{6} \times 60 = 40^{\circ}$

$\displaystyle \therefore A = 60^{\circ} - 40^{\circ} = 20^{\circ} B = 60^{\circ} C = 60^{\circ} + 40^{\circ} = 100^{\circ}$

In radians

$\displaystyle A = 20 \times \frac{\pi}{180} = \Big( \frac{\pi}{9} \Big)^c B = 60 \times \frac{\pi}{180} = \Big( \frac{\pi}{3} \Big)^c C = 100 \times \frac{\pi}{180} = \Big( \frac{5\pi}{9} \Big)^c$

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Question 10: The number of sides of two regular polygons are as $\displaystyle 5:4$ and the difference between their angles is 9^{\circ}. Find the number of sides of the polygon.

Answer:

$\displaystyle \text{Let } n \text{ and } m$ be the number of sides of two regular polygons respectively.

$\displaystyle \therefore \frac{n}{m} = \frac{5}{4} \Rightarrow n = \Big( \frac{5}{4} \Big) m$

Also given

$\displaystyle \Big( \frac{2n-4}{n} \Big) \times 90^{\circ} - \Big( \frac{2m-4}{m} \Big) \times 90^{\circ} = 9^{\circ}$

$\displaystyle \Rightarrow \frac{2n-4}{n} - \frac{2m-4}{m} = \frac{1}{10}$

$\displaystyle \Rightarrow \frac{(2n-4)m - (2m-4)n}{mn} = \frac{1}{10}$

$\displaystyle \Rightarrow \frac{2mn - 4m - 2mn + 4n}{mn} = \frac{1}{10}$

$\displaystyle \Rightarrow \frac{4n - 4m}{mn} = \frac{1}{10}$

$\displaystyle \text{Substituting } n = \Big( \frac{5}{4} \Big) m$

$\displaystyle \Rightarrow \frac{4 ( \frac{5}{4} ) m - 4m}{m ( \frac{5}{4} ) m} = \frac{1}{10} \Rightarrow \frac{4m}{5m^2} = \frac{1}{10} \Rightarrow \frac{4}{5m} = \frac{1}{10} \Rightarrow m = 8$

$\displaystyle \therefore n = \frac{5}{4} \times 8 = 10$

Hence $\displaystyle n = 10 \text{ and } m = 8$

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Question 11: A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by $\displaystyle 25^{\circ}$ in a distance of $\displaystyle 40 \text{ meters }$

Answer:

$\displaystyle \text{Let } AB$ be the rail road

$\displaystyle \therefore \angle AOB = 25^{\circ} = 25 \times \frac{\pi}{180} = \Big( \frac{5\pi}{36} \Big)^c$

$\displaystyle \text{We know that } \theta = \frac{\text{arc} }{\text{radius} }$

$\displaystyle \Rightarrow \frac{5\pi}{36} = \frac{40}{r}$

$\displaystyle \Rightarrow r = \frac{36 \times 40}{5 \pi} = \frac{288}{\pi} \text{ meters }$

$\displaystyle \Rightarrow r = 91.64 \text{ meters }$

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Question 12: Find the length which is at a distance $\displaystyle 5280$ m will subtend an angle of $\displaystyle 1'$ at the eye.

Answer:

$\displaystyle \angle AOB = 1'$

$\displaystyle \text{Given Radius } = 5280 \text{ meters }$

$\displaystyle \angle AOB = \Big( \frac{1}{60} \times \frac{\pi}{180} \Big)^c$

$\displaystyle \text{We know that } \theta = \frac{\text{arc} }{\text{radius} }$

$\displaystyle \Rightarrow \frac{1}{60} \times \frac{\pi}{180} = \frac{l}{5280}$

$\displaystyle \Rightarrow l = \frac{5280}{60 \times 180} \times \frac{22}{7} = 1.5365 \text{ meters }$

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Question 13: A wheel makes $\displaystyle 360$ revolutions in a minute. Through how many radians does it turn in $\displaystyle 1$ second.

Answer:

$\displaystyle \text{Given } 360$ revolutions per minute $\displaystyle \Rightarrow 6$ revolution per second

$\displaystyle \text{In } 1$ revolution, wheel will cover $\displaystyle 360^{\circ}$

$\displaystyle \therefore$ In one second the wheel will cover $\displaystyle 360 \times 6 = 2160^{\circ}$

$\displaystyle \therefore \text{ In radians, } 2160 \times \frac{\pi}{180} =(12\pi)^c$

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Question 14: Find the angle in radians through which a pendulum swings if its length is $\displaystyle 75 \text{ cm }$ and the tip describes as arc of length $\displaystyle \text{(i) } 10 \text{ cm } \text{(ii) } 15 \text{ cm } \text{(iii) } 21 \text{ cm }$

Answer:

$\displaystyle \text{(i) Let } OA =$ length of pendulum $\displaystyle = 75 \text{ cm } = 0.75$ m

$\displaystyle AB = \text{arc} \ AB = 10 \text{ cm } = 0.1$ m

$\displaystyle \therefore \theta = \frac{\text{arc} }{\text{radius} } = \frac{0.1}{0.75} = \Big( \frac{2}{15} \Big)^c$

$\displaystyle \text{(ii) } OB = 0.75$ m

$\displaystyle AB = 15 \text{ cm } = 0.15$ m

$\displaystyle \therefore \theta = \frac{\text{arc} }{\text{radius} } = \frac{0.15}{0.75} = \Big( \frac{1}{5} \Big)^c$

$\displaystyle \text{(iii) } OB = 0.75$ m

$\displaystyle AB = 21 \text{ cm } = 0.21 \text{ cm }$

$\displaystyle \therefore \theta = \frac{\text{arc} }{\text{radius} } = \frac{0.21}{0.75} = \Big( \frac{7}{25} \Big)^c$

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Question 15: The radius of a circle is $\displaystyle 30 \text{ cm.}$ Find the length of an arc of this circle, if the length of the chord of the arc is $\displaystyle 30 \text{ cm.}$

Answer:

$\displaystyle OA = OB =$ radius of circle $\displaystyle = 30 \text{ cm } = 0.3$ m

$\displaystyle AB =$ chord $\displaystyle AB = 30 \text{ cm } = 03$ m

$\displaystyle \therefore \triangle ABC$ is equilateral triangle.

$\displaystyle \text{Let } \text{arc} AB = l$

$\displaystyle \Rightarrow \theta = 60^{\circ} = \Big( \frac{\pi}{3} \Big)^c$

$\displaystyle \therefore \frac{\pi}{3} = \frac{l}{0.3}$

$\displaystyle \Rightarrow 0.1 \pi$ m $\displaystyle = 10 \pi \text{ cm }$

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Question 16: A railway train is travelling on a circular curve of $\displaystyle 1500$ m radius at a rate of $\displaystyle 66$ km/hr. Through what angle has it turned in $\displaystyle 10$ seconds?

Answer:

$\displaystyle OA = OB =$ radius $\displaystyle = 1500$ m

The train turns $\displaystyle \angle AOB = \theta \text{In } 10$ seconds

$\displaystyle \text{Speed of train } = 66 \text{ km/hr } = \frac{66 \times 1000}{3600} \text{ m/s } = \frac{110}{6} \text{ m/s }$

$\displaystyle \text{In } 10 \text{ seconds, train travels } = \frac{110}{6} \times 10 = \frac{1100}{6} \text{ m }$

$\displaystyle \therefore \text{arc} AB = \frac{1100}{6}$ m

$\displaystyle \therefore \theta = \frac{\text{arc} }{\text{radius} } = \frac{1100}{6 \times 1500} = \Big( \frac{11}{90} \Big)^c$

$\displaystyle \therefore \text{ the train will turn by } \Big( \frac{11}{90} \Big)^c \text{In } 10 \text{ seconds }$

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Question 17: Find the distance from the eye at which a coin of $\displaystyle 2 \text{ cm }$ diameter should be held so as to conceal the full moon whose angular diameter is $\displaystyle 31'$ .

Answer:

$\displaystyle \text{Let } r$ be the distance at which coin needs to be placed to completely conceal the full moon.

$\displaystyle \therefore \theta = 31' = \Big( \frac{31}{60} \Big)^{\circ} = \Big( \frac{31}{60} \times \frac{\pi}{180} \Big)^c$

$\displaystyle \text{Given } \text{arc} \ AB = 2 \text{ cm } = 0.02 \text{ cm }$

$\displaystyle \therefore \theta = \frac{\text{arc} }{\text{radius} }$

$\displaystyle \Rightarrow \frac{31}{60} \times \frac{\pi}{180} = \frac{0.02}{r}$

$\displaystyle \Rightarrow r = \frac{60 \times 180 \times 0.02}{31 \times \pi} = 2.217$ m

Therefore the coin should be placed $\displaystyle 2.217$ m away from the eye.

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Question 18: Find the diameter of the sun in km supposing that it subtends an angle of $\displaystyle 32'$ at the eye of an observer. Given that the distance of the sun is $\displaystyle 91 \times 10^6$ km.

Answer:

$\displaystyle \text{Given } \angle AOB = 32' = \Big( \frac{32}{60} \Big)^{\circ} = \Big( \frac{32}{60} \times \frac{\pi}{180} \Big)^c$

$\displaystyle \text{Since } \theta = \frac{\text{arc} }{\text{radius} }$

$\displaystyle \Rightarrow \frac{32}{60} \times \frac{\pi}{180} = \frac{AB}{91 \times 10^6}$

$\displaystyle \Rightarrow AB = \frac{32 \times \pi \times 91 \times 10^6}{60 \times 180} = 8.47 \times 10^5$ km

Therefore Distance of sun is $\displaystyle 847407.4$ km

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Question 19: If the arcs of the same length in two circles subtend angles $\displaystyle 65^{\circ} \text{ and } 110^{\circ}$ at the center, find the ratio of their radii.

Answer:

$\displaystyle \text{Let } C_1 \text{ and } C_2$ be two circles with same arc length $\displaystyle l$

$\displaystyle \text{If } \theta_1 \text{ and } \theta_2$ are two angles subtended by the arcs on respective circles

$\displaystyle \therefore \theta_1 = 65^{\circ} = \Big( \frac{65\pi}{180} \Big)^c$

$\displaystyle \text{and } \theta_2 = 110^{\circ} = \Big( \frac{110\pi}{180} \Big)^c$

$\displaystyle \text{We know, } \theta = \frac{\text{arc} }{\text{radius} }$

$\displaystyle \therefore \theta_1 = \frac{l}{r} \Rightarrow r = \frac{l}{\theta_1}$

$\displaystyle \text{Similarly, } \theta_2 = \frac{l}{R} \Rightarrow R = \frac{l}{\theta_2}$

$\displaystyle \text{Therefore } \frac{r}{R} = \frac{\frac{l}{\theta_1}}{\frac{l}{\theta_2}} = \frac{\theta_2}{\theta_1} = \frac{\frac{110\pi}{180}}{ \frac{65\pi}{180}} = \frac{22}{13}$

$\displaystyle \therefore r: R = 22:13$

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Question 20: Find the degree measure of the angle subtended at the center of the circle of radius $\displaystyle 100 \text{ cm }$ by an arc of length $\displaystyle 22 \text{ cm.}$ (Use $\displaystyle \pi = \frac{22}{7}$ ):