# NCRT Solutions Excercise 1

**Question 1**: Find the radian measures corresponding to the following degree measures:

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

**Answer 1**: (i) 25°

As we know 180° = π radian

∴25^{o}= π /180 x 25 radian=5 π/36 radian

(ii) –47° 30'

–47° 30'

=-95/2 degree

Since 180° = π radian

=95/2^{o} = π/180 x (-95/2) radian=(-19/36x2) π radian=-19/72 π radian

∴-47^{o} 30’=-19/72 π radian

(iii) 240°

We know that 180° = π radian

∴520^{o}= π /180 x 520 radian= 26π /9 radian

**Question 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

**Answer 2:** (i) 11/16

As we know that π radian = 180°

Therefore 11/16 radian=180/ π x 11/16 degree

= 45 x 11/ π x 4 degree

=45 x 11 x 7/ 22 x 4 degree

=315/8 degree

=39^{o}+3 x 60/8 minutes [1^{o}=60’]

=39^{o} + 22’ + ½ minutes

=39^{o}22’30” [1’=60]

(ii) – 4

We know that π radian = 180°

-4 radian=180/π x (-4) degree=180 x 7(-4)/22 degree

=-2520/11 degree=

=-229^{o}+1 x 60/11 minutes [1^{o}=60’]

=-229^{o}+5’+5/11 minutes

=-229^{o}5’27” [1’=60”]

(iii) 5 π/3

We know that π radian = 180°

5 π/3 radian=180/π x 5 π/3 degree=300^{o}

(iv) 7 π /6

We know that π radian = 180°

∴ 7 π /6 radian=180/π x 7 π/6 =210^{o}

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

**Answer 3:** Number of revolutions made by the wheel in 1 minute = 360

∴ Number of revolutions made by the wheel in 1 second =360/60=6

In one complete revolution, the wheel turns through an angle of 2π radian.

Hence, in 6 complete revolutions, it will turn through an angle of 6 × 2π radian, i.e., 12 π radian

Thus, in one second, the wheel turns an angle of 12π radian.

**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.

(Use π =22/7)

**Answer 4:** We know that 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

Therefore, for= 100 cm, l = 22 cm, we have

θ =22/100 radian=180/π x 22/100 degree=180 x 7 x 22/22 x 100 degree

=126/10 degree=12^{o}36’ [1^{o}=60]

Thus, the required angle is 12°36′.

**Question 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.

**Answer 5:** Diameter of the circle = 40 cm

∴Radius (r) of the circle =40/2 cm=20 cm

Let AB be a chord (length = 20 cm) of the circle.

In âˆ†OAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, âˆ†OAB is an equilateral triangle.

∴ θ = 60° = π /3 radian

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

θ =l/r

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

**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.

**Answer 6:** Let the radii of the two circles be r_{1} and r_{2}. Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r_{1}, and let an arc of length l subtend an angle of 75° at the centre of the other circle of radius r_{2}.

Now, 60°= π/3 radian and 75° =5 π/12 radian

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

θ =1/r or l=rθ

∴l=r_{1}π /3 and l=r_{2}5π /12

⇒ r_{1}π /3= r_{2}5π /12

⇒r_{1}= r_{2}5 /4

⇒r_{1}/r_{2}=5/4

Thus, the ratio of the radii is 5:4.

**Question 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

**Answer 7**: We know that in a circle of radius r unit, if an arc of length l unit subtends

an angle θ radian at the centre, then θ =l/r

It is given that r = 75 cm

(i) Here, l = 10 cm

θ =10/75 radian=2/15 radian

(ii) Here, l = 15 cm

θ =15/75 radian =1/5 radian

(iii) Here, l = 21 cm

θ =21/75 radian=7/25 radian