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

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

Answer:

i) \Big( \frac{9\pi}{5} \Big)^c = \Big( \frac{9\pi}{5} \times \frac{180}{\pi} \Big)^o = 324^o

ii) \Big( \frac{-5\pi}{6} \Big)^c = \Big( \frac{-5\pi}{6} \times \frac{180}{\pi} \Big)^o = -150^o

iii) \Big( \frac{18\pi}{5} \Big)^c = \Big( \frac{18\pi}{5} \times \frac{180}{\pi} \Big)^o = 648^o

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

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

v) (11)^c =\Big( 11 \times \frac{180}{22} \times 7  \Big)^o = 630^o

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

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

i) 300^o    ii) 35^o    iii) -56^o    iv) 135^o    v) -300^o    vi) 7^o 30'    vi) 125^o 30'    vii) -47^o 30'

Answer:

i) 300^o = \Big( 300 \times \frac{\pi}{180}   \Big)^c = \Big( \frac{5\pi}{3}   \Big)^c

ii) 35^o = \Big( 35 \times \frac{\pi}{180}   \Big)^c = \Big( \frac{7\pi}{36}   \Big)^c

iii) -56^o = \Big( -56 \times \frac{\pi}{180}   \Big)^c = \Big( \frac{-14\pi}{45}   \Big)^c

iv) 135^o = \Big( 135 \times \frac{\pi}{180}   \Big)^c = \Big( \frac{3\pi}{4}   \Big)^c

v) -300^o = \Big( -300 \times \frac{\pi}{180} \Big)^c = \Big( \frac{-5\pi}{3}   \Big)^c

vi) 7^o30' = \Big( 7 \frac{30}{60} \Big)^o = \Big(7 \frac{1}{2} \Big)^o = \Big( \frac{15}{2} \times \frac{\pi}{180} \Big)^c = \Big( \frac{\pi}{24} \Big)^c 

vii) 125^o30' = \Big( 125 \frac{30}{60} \Big)^o = \Big(125 \frac{1}{2} \Big)^o = \Big( \frac{251}{2} \times \frac{\pi}{180} \Big)^c = \Big( \frac{251\pi}{360} \Big)^c 

viii) -47^o30' = \Big( -47 \frac{30}{60} \Big)^o = \Big(-47 \frac{1}{2} \Big)^o = \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 \frac{2 \pi}{5} radians. Express the angles in degrees.

Answer:

Let x_1  and x_2  be two acute angles of a right angles triangle.

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

Also in a right angled triangle

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

Adding i) and ii) we get

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

\Rightarrow 2 x_1 = \frac{9}{10} \pi 

\Rightarrow x_1 = \frac{9}{20} \pi = \frac{9}{20} \times 180 = 81^o 

\therefore x_2 = 90- 81 = 9^o 

Hence x_1 = 81^o  and x_2 = 9^o

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

Answer: Let the three angles be x_1, x_2 and x_3

\therefore x_1 = \frac{2}{3} x gradiant = \Big(  \frac{2}{3} x \times \frac{90}{100} \Big)^o = \Big( \frac{3}{5} x  \Big)^o

x_2 = \Big( \frac{3}{2} x  \Big)^o

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

In the triangle

x_1 + x_2 + x_3 = \pi

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

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

\Rightarrow \frac{45x}{10} = \pi    \Rightarrow x = \frac{2}{9} \pi = 40^o

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

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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 n sides = \Big( \frac{2n-4}{n} \Big) \times 90^o

i) Pentagon

Interior angle = \Big( \frac{2 \times 5-4}{5} \Big) \times 90^o = 108^o

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

ii) Octagon

Interior angle = \Big( \frac{2 \times 8-4}{8} \Big) \times 90^o = 135^o

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

iii) Heptagon

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

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

iv) Duodecagon

Interior angle = \Big( \frac{2 \times 12-4}{12} \Big) \times 90^o = 150^o

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 120^o . Express the angles in radians.

Answer:

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

We know, sum of angles in quadrilateral = 360^o

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

\Rightarrow 4a = 360    \Rightarrow 4 = 90^o

Largest angle a+3d = 120

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

\therefore angles are 60^o, 80^o, 100^o and 120^o

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

\Rightarrow \frac{\pi}{3} , \frac{4\pi}{9} , \frac{5\pi}{9} ,  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 1:120 . Find the angles in radians.

Answer:

Let A, B and C be the three angles

It is given that A, B and C are in an AP

Let A = a-d, \ B = a , and C = a+d

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

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

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

\Rightarrow 120 - 120d = 1

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

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

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

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

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 3:2 and the number of sides in first is twice that in second. Determine the number of sides of two polygons.

Answer:

Let n and m be the number of sides of two regular polygons respectively.

\therefore \frac{\frac{2n-4}{n} \times 90^o}{\frac{2m-4}{m} \times 90^o} = \frac{3}{2}

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

Also given n = 2m

Substituting in i) we get

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

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

\Rightarrow 4m - 4 = 6m - 12

\Rightarrow 2m = 8 \Rightarrow m = 4

\therefore n = 2 \times 4 = 8

Hence n = 8 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:

Let A, B and C be the three angles

Let A = a - d, B = a and C = a+d

\therefore a-d + a + a + d = 180^o

\Rightarrow 3a = 180^o     \Rightarrow a = 60^o

Given C = 5 A

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

\therefore d = \frac{4}{6} \times 60 = 40^o

\therefore A = 60^o - 40^o = 20^o     B = 60^o     C = 60^o + 40^o = 100^o

In radians

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 5:4 and the difference between their angles is 9^o. Find the number of sides of the polygon.

Answer:

Let n and m be the number of sides of two regular polygons respectively.

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

Also given

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

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

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

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

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

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

\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

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

Hence n = 10 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 25^o in a distance of 40 meters.

Answer:

Let AB be the rail road

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

We know that \theta = \frac{arc}{radius}

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

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

\Rightarrow r = 91.64 meters

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

Answer:

\angle AOB = 1'

Given Radius = 5280 meters

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

We know that \theta = \frac{arc}{radius}

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

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

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

Answer:

Given 360 revolutions per minute \Rightarrow 6 revolution per second

In 1 revolution, wheel will cover 360^o

\therefore In one second the wheel will cover 360 \times 6 = 2160^o

\therefore 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 75 cm and the tip describes as arc of length i) 10 cm   ii) 15 cm   iii) 21 cm

Answer:

i) Let OA = length of pendulum = 75 cm = 0.75 m

AB = arc \ AB = 10 cm = 0.1 m

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

ii) OB = 0.75 m

AB = 15 cm = 0.15 m

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

iii) OB = 0.75 m

AB = 21 cm = 0.21 cm

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

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Question 15: The radius of a circle is 30 cm. Find the length of an arc of this circle, if the length of the chord of the arc is 30 cm.

Answer:

OA = OB = radius of circle = 30 cm = 0.3 m

AB = chord AB = 30 cm = 03 m

\therefore \triangle ABC is equilateral triangle.

Let arc AB = l

\Rightarrow \theta = 60^o = \Big(  \frac{\pi}{3}  \Big)^c 

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

\Rightarrow 0.1 \pi m = 10 \pi cm

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

Answer:

OA = OB = radius = 1500 m

The train turns \angle AOB = \theta in 10 seconds

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

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

\therefore arc AB = \frac{1100}{6}  m

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

\therefore the train will turn by \Big( \frac{11}{90}  \Big)^c in 10 seconds

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

Answer:

Let r be the distance at which coin needs to be placed to completely conceal the full moon.

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

Given arc \ AB = 2 cm = 0.02 cm

\therefore \theta = \frac{arc}{radius}

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

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

Therefore the coin should be placed 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 32' at the eye of an observer. Given that the distance of the sun is 91 \times 10^6 km.

Answer:

Given \angle AOB = 32' = \Big( \frac{32}{60} \Big)^o = \Big( \frac{32}{60} \times \frac{\pi}{180} \Big)^c

Since \theta = \frac{arc}{radius}

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

\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 847407.4 km

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Question 19: If the arcs of the same length in two circles subtend angles 65^o and 110^o at the center, find the ratio of their radii.

Answer:

Let C_1 and C_2 be two circles with same arc length l

If \theta_1 and \theta_2 are two angles subtended by the arcs on respective circles

\therefore \theta_1 = 65^o = \Big( \frac{65\pi}{180} \Big)^c

and \theta_2 = 110^o = \Big( \frac{110\pi}{180} \Big)^c

We know, \theta = \frac{arc}{radius}

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

Similarly, \theta_2 = \frac{l}{R} \Rightarrow R = \frac{l}{\theta_2}

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}

\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 100 cm by an arc of length 22 cm. (Use \pi = \frac{22}{7} ):

Answer:

Let AB = arc AB = 22  cm

OA = OB = r = 100 cm

Let \theta be the angle subtended by the arc \ AB at center O

\therefore \theta = \frac{arc}{radius}

\theta = \Big( \frac{22}{100} \Big)^c     \Rightarrow \Big( \frac{22}{100} \times \frac{180}{\pi} \Big)^o = 12^o36'