Ex 3.1 Trigonometric function ncert solution class 11

 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

 

https://gmath.in/chapter-2-miscellaneous-sets-ncert-maths-solution-class-11/

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