EXERCISE 3.1(Trigonometric function)

1. Find the radian measures corresponding to the following degree measures:(Ex 3.1 Trigonometric function ncert solution class 11)

(i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°

Solution: (i) $25^{\circ}$

Since, $1^{\circ}= \frac{\pi}{180}$ Radian

Then,

$25^{\circ}=\frac{\pi}{180}\times 25$ $= \frac{5 \pi}{36}$

(ii) $-47^{\circ}30'$

$1^{\circ}=60'$

And $1' = \frac{1}{60}$ degree

Then, $-47^{\circ}30'= -(47+\frac{30}{60})$

$= -47\frac{1}{2}= -\frac{95}{2}$ degree

Since, $1^{\circ}= \frac{\pi}{180}$ Radian

Then,

$-(\frac{95}{2})^{\circ}=\frac{\pi}{180}\times -\frac{95}{2}$

$= -\frac{19 \pi}{72}$ Radian

(iii) $240^{\circ}$

Since, $1^{\circ}= \frac{\pi}{180}$ Radian

Then,

$240^{\circ}=\frac{\pi}{180}\times 240$

$= \frac{4 \pi}{3}$ radian

(iv) $520^{\circ}$

Since, $1^{\circ}= \frac{\pi}{180}$ radian

Then,

$520^{\circ}=\frac{\pi}{180}\times 520$

$= \frac{26 \pi}{9}$ radian

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7)

(i) 11/16

(ii) -4

(iii) 5π/3

(iv) 7π/6

Solution: (i) $\frac{11}{16}$

$1 \text{ radian} = \frac{180}{\pi}\text{ degree}$

Then

$\frac{11}{16} \text{ radian}= \frac{180}{\pi}\times \frac{11}{16}$

$= \frac{180\times 11\times 7}{22\times 16}$

$= \frac{315}{8}$ degree

$\Rightarrow \frac{315}{8}= 39^{\circ}+ \frac{3\times 60}{8}$ minutes

$= 39^{\circ}+22'+\frac{1\times 60}{2}$ second

$= 39^{\circ}22' 30''$

Hence,

$\frac{11}{16} \text{ radian}=39^{\circ}22' 30''$

(ii) $-4$

$1 radian = \frac{180}{\pi}$

$-4 \text{ radian} = \frac{180}{\pi}\times (-4)\text{ degree}$

$= -\frac{180\times 4\times 7}{22}$

$= -\frac{2520}{11}= -229\frac{1}{11}\text{ degree}$

Since,

$-\frac{2520}{11}\text{ radian}= -(229^{\circ}+\frac{1\times 60}{11}\text{ minutes})$

$= -(229^{\circ}+5'+\frac{5\times 60}{11}\text{ second})$

$= -(229^{\circ}+5'+ 27'')$

$-4 \text{ radian}= -229^{\circ}5' 27''$

(iii) $\frac{5\pi}{3}$

$1 \text{ radian} = \frac{180}{\pi}$

$\frac{5\pi}{3}\text{ radian} = \frac{180}{\pi}\times \frac{5\pi}{3} = 300^{\circ}$

(iv) $\frac{7\pi}{6}$

$1 \text{ radian} = \frac{180}{\pi}$

$\frac{7\pi}{6}\text{ radian} = \frac{180}{\pi}\times \frac{7\pi}{6} = 210^{\circ}$

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

Solution: It is given that

No. of revolutions made by the wheel in 1 minute = 360

No. of revolution in 1 second = 360/60 = 6

We know that

The wheel turns an angle of 2π radian in one complete revolution.

In 6 complete revolutions = 6 × 2π radian = 12 π radian

Therefore, in one second, the wheel turns an angle = 12π radian.

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 (Use π = 22/7).

Solution: Let radius of circle = r

and arc subtends an angle at the centre $= \theta$

$\theta = l/r$

Since, $r = 100$ cm, $l = 22$ cm

$\theta = \frac{22}{100}$ radian

$= \frac{180}{\pi}\times \frac{22}{100}$

$= \frac{180\times 22 \times 7}{22 \times 100}$

$= \frac{126}{10} = \frac{63}{5}$ degree

$= 12^{\circ}+\frac{3\times 60}{5}\text{ minutes}$

$= 12^{\circ}36'$

Therefore the required angle is $12^{\circ}36'$

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

Solution: The dimensions of the circle are

Diameter = 40 cm

Radius = 40/2 = 20 cm

Consider AB be as the chord of the circle i.e. length = 20 cm

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 10

In ΔOAB,

Radius of circle = OA = OB = 20 cm

Given AB = 20 cm

Hence, ΔOAB is an equilateral triangle.

θ = 60° = π/3 radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre

We get θ = l/r

NCERT Solutions for Class 11 Chapter 3 Ex 3.1 Image 11

Therefore, the length of the minor arc of the chord is 20π/3 cm.

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 $r_1$ and $r_2$ are radius of two circles.

And arc length l subtend angle in first circle is 60° at the centre and arc length l subtend an angle 75° at the centre

Since, $60^{\circ} = \pi/3$ radian

$75^{\circ}=5\pi/12$ radian

We know that

$l = r_1\theta_1 = \frac{\pi}{3}r_1$

And $l = r_2\theta_2 = \frac{5\pi}{12}r_2$

Hence, $\frac{\pi}{3}r_1= \frac{5\pi}{12}r_2$

$\frac{r_1}{r_2}=\frac{5}{4}$

Therefore ratio between the radius = 5:4

7. Find the angle in radian though 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 a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then θ = 1/r

We know that r = 75 cm

(i) l = 10 cm

We know that θ = l/r

θ = 10/75 radian

By further simplification

θ = 2/15 radian

(ii) l = 15 cm

We know that θ = l/r

θ = 15/75 radian

⇒ θ = 1/5 radian

(iii) l = 21 cm

We know that θ = l/r

θ = 21/75 radian

⇒ θ = 7/25 radian

Chapter 2 Miscellaneous sets ncert maths solution class 11

EXERCISE 3.1(Trigonometric function)

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