11th NCERT/CBSE Trigonometric Functions Exercise 3.1 Questions 7
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Question (1)

Find the radian measures corresponding to the following degree measures
(i) 25°   (ii) -47°37'   (iii) 240°   (iv) 52°

Solution

(i) 25°
We know that 180° = π radian ${25^o} = \frac{\pi }{{180}} \times 25radian = \frac{{5\pi }}{{36}}radian$ (ii) -47° 30’
$- {47^o}30' = - 47\frac{1}{2}\text{degree} \;[{1^0} = 60']$ $= \frac{{ - 95}}{2}\text{degree}$ Since 180° = π radian
$\frac{{ - 95}}{2}\text{degree} = \frac{\pi }{{180}} \times \left( {\frac{{ - 95}}{2}} \right) \text{radian}$ $\frac{{ - 95}}{2}\text{degree} = \frac{{ - 19}}{{72}}\pi \; text{radian}$ $- {47^o}30' = \frac{{ - 19}}{{72}}\pi \text{radian}$
(iii) 240°
We know that 1° = $\frac{\pi }{{180}} \; \text{radian}$ ${180^o} = 240 \times \frac{\pi }{{180}} = \frac{4}{3}\pi \; radian$ (iv) 520°
We know that 1° = $\frac{\pi }{{180}} \; \text{radian}$ ${180^o} = 520 \times \frac{\pi }{{180}} = \frac{26}{9}\pi \; radian$

Question (2)

Find the degree measures corresponding to the following radian measures $\left( {\text{Use} \; \; \pi = \frac{{22}}{7}} \right)$
$\left( i \right)\;\;\frac{{11}}{{16}}$     (ii) -4
$\left( {iii} \right)\;\;\frac{{5\pi }}{3}$     $\left( {iv} \right)\;\;\frac{{7\pi }}{6}$

Solution

(i) $\left( i \right)\;\;\frac{{11}}{{16}}$
Since π radian = 180° =
$\frac{{11}}{{16}}radian = \frac{{180}}{\pi } \times \frac{{11}}{{16}}\text{degree}$ $\frac{{11}}{{16}}radian = \frac{{180}}{{\frac{{22}}{7}}} \times \frac{{11}}{{16}}\text{degree}$ $\frac{{11}}{{16}}radian = \frac{{\cancel{180}^{45} \times 7}}{\cancel{{22}_{2}}} \times \frac{\require{cancel} \cancel{{11}}}{\cancel{{16}}_4}\text{degree}$ $\frac{{11}}{{16}}radian = \frac{{45 \times 7}}{2} \times \frac{1}{4}\text{degree}$ $\frac{{11}}{{16}}radian = \frac{{315}}{8}\text{degree}$ $\frac{{11}}{{16}}radian = \frac{{312 + 3}}{8}\text{degree}$ $\frac{{11}}{{16}}radian = \left( {39 + \frac{3}{8}} \right)\text{degree}$ 39° ---(1)
Now 1°= 60' (minutes)
$\frac{3}{8}\text{degree} = \left( {\frac{{45}}{2}} \right)\text{minutes} = \left( {22 + \frac{1}{2}} \right)\text{minutes}$ 39° and 22'
Now 1' = 60 second
$\frac{1}{2}\text{minutes} = \frac{1}{2} \times 60 = 30\text{second}$ $\frac{{11}}{{16}}radian = {39^o}22'30"$
(ii) -4
Since π radian = 180° =
$- 4radian = - \frac{{4 \times 180}}{\pi } = - \frac{{4 \times 180 \times 7}}{{22}}\text{degree}$ $- 4radian = \frac{{5040}}{{22}} = \frac{{5038 + 2}}{{22}} = 229 + \frac{2}{{22}}\text{degree}$ -229° ....(1)
Now 1° = 60'(minutes)
$\frac{2}{{22}}\deg ree = \frac{2}{{22}} \times 60 = \frac{{60}}{{11}}\min utes$ $\frac{{60}}{{11}}\min utes = \frac{{55 + 5}}{{11}} = \left( {5 + \frac{5}{{11}}} \right)\min utes$ 5 minutes ....(2) Now 1 minutes = 60 second
$\frac{5}{{11}}\min utes = \frac{5}{{11}} \times 60\sec$ $\frac{5}{{11}}\min utes = \frac{{300}}{{11}} = \frac{{297 + 3}}{{11}} = \left( {27 + \frac{3}{{11}}} \right)\sec$ 27 sec ....(3) -4 radiants= -229° 5' 27"
$\left( {iii} \right)\;\;\frac{{5\pi }}{3}$ Since π radian = 180°
$\frac{{5\pi }}{3}radian = \frac{{180}}{\pi } \times \frac{{5\pi }}{3}\text{degree} = {300^o}$
$\left( {iv} \right)\;\;\frac{{7\pi }}{6}$
$\frac{{7\pi }}{6}radian = \frac{{180}}{\pi } \times \frac{{7\pi }}{6}\text{degree} = {210^o}$

Question (3)

A wheel makes 360 revolutions in one minute. Through how many radians does in turn in one second?

Solution

Number of revolution of wheal in minute = 360
Number of revolution per second = 360/60 = 6 revolution per second Now one revolution = 2π radiant
∴ radiants in one second = 6× 2π = 12π

Question (4)

Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm .
$\left( {\text{Use} \; \; \pi = \frac{{22}}{7}} \right)$

Solution

In circle of radius r, Angle subtended by a arch of length l is &theta = l/r
Given r= 100 cm and l = 22 cm
∴ θ = 22/100 radian We know π radian = 180° $\frac{{22}}{{100}}radian = \frac{{180}}{\pi } \times \frac{{22}}{{100}}\deg ree$ $\frac{{22}}{{100}}radian = \frac{{180}}{{22}} \times 7 \times \frac{{22}}{{100}} = \frac{{126}}{{10}}\deg ree$ $\frac{{22}}{{100}}radian = \frac{{120 + 6}}{{10}} = \left( {12 + \frac{6}{{10}}} \right)\deg ree$ 12° ...(1)
Now 1° = 60minutes
$\frac{6}{{10}}\deg ree = \frac{6}{{10}} \times 60 = 36\min utes$ 36 minutes ....(2) Angle θ = 12° 36'

Question (5)

In a circle of diameter 40cm, the length of chord is 20cm. Find the length of minor arc of the chord

Solution

Dameter of the circle 40cm, or radius is 20 cm ,length of the chord =20 cm
Thus angle formed by chord and two sides of trabgle formed at centre is 60 as shown in figure as it form a equilateral triangle

In circle of radius r, Angle subtended by a arch of length l is θ = l/r
$\frac{\pi }{3} = \frac{l}{{20}}$ $l = \frac{{20}}{3}\pi$ Now l is length of minor arc $\overset{ \huge\frown}{AB} = \frac{{20}}{3}\pi$

Question (6)

If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii

Solution

Let the radii of the two circles be r1 and r2. Let an arc of length l subtends an angle of 60° at the circle of radius r1 br/> While let an arc of length l subtend an angle of 75° at the centre of the circle of radius r2 Now 60° = π/3 radiant and 75°= 5π/12 radiant
In circle of radius r, Angle subtended by a arch of length l is θ = l/r or l = r θ
$l = \frac{{{r_1}\pi }}{3}\, \text{and} \; l = \frac{{{r_2}5\pi }}{{12}}$ $\Rightarrow \frac{{{r_1}\pi }}{3} = \frac{{{r_2}5\pi }}{{12}}$ $\Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{5}{4}$ ∴ Ratio of radii is 5:4

Question (7)

Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm   (ii) 15 cm   (iii) 21 cm

Solution

In circle of radius r, Angle subtended by a arch of length l is θ = l/r or l = rθ
Given r = 75 cm
(i) given l=10 cm
$\theta = \frac{{10}}{{75}} = \frac{2}{{15}}radian$ (ii) given l =15 cm $\theta = \frac{{15}}{{75}} = \frac{1}{5}radian$ (iii) given 21 cm $\theta = \frac{{21}}{{75}} = \frac{7}{{25}}radian$