Trigonometric Functions
NCERT Solution
Exercise 1
Question 1: Find the radian measures corresponding to the following degree measures:
(i) 25° (ii) -47°30’ (iii) 240° (520°
Solution:We know, π radian = 180°
So, `1°=(π)/(180°)`
`25°=25xx(π)/(180°)`
`=(5π)/(36)` radian
`-47°30’=-(95)/2`
So, `(95)/2=(95)/(2)xx(π)/(180)`
`=-(19π)/(72)` radian
`240°=240xx(π)/(180)`
`=(4π)/3` radian
`520°=520xx(π)/(180)`
`=(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`
Solution: We know that π radian = 18°
Therefore, 1 radian `=(180°)/(π)°`
(i) `(11)/(16)` radian `= (11)/(16)xx(180)/(π)`
`=(11)/(16)xx(180)/(22)xx7`
`=1/4xx(45)/(2)xx7=(315)/(8)°`
`=39°22’30’’`
(ii) -4 radian `=-(4xx(180°)/π)`
`=-4xx(180)/(22)xx7=-(2520)/(11)`
`=-229°5’27’’`
(iii) `(5π)/(3)` radian `=(5π)/(3)xx(180)/(π)`
`=5xx60=300°`
(iv) `(7π)/(6)` radian `=(7π)/(6)xx(180)/(π)`
`=7xx30=210°`
Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Solution: We know that 1 minute = 60 second
Since, in 60 sec number of revolutions = 360
Hence, in 1 sec number of revolutions `=(360)/(60)=6`
In one revolution, the wheel makes angle of 360°
So, in 6 revolutions angle made by wheel `=360°xx6`
`=2xx180°xx6=2xxπxx6` radian
`=12 π` 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)
Solution: Given, length of arc, l = 22cm
Radius of the circle, r = 100 cm
We know that, l = r θ
Therefore, `θ=l/r`
Where, l = length of arc
r = radius of circle
ѳ = angle subtended at the centre.
Therefore,
`θ=(22)/(100)` radian
Or, `θ=0.22` radian
Now, 0.22 radian `=0.22xx((180)/(π))°`
`=(0.22xx(180xx7)/(22))°`
`=((180xx7)/(100))°`
`=12.6°=12°36’`
Question 5: In a circle of diameter 40 cm, the length of cord is 20cm. Find the length of minor arc of the chord.
Solution:
Let ABC is a circle with centre O.
The diameter AB of the circle = 40cm
The BC is the chord of length = 20cm
Therefore, radius OB = 20cm
The length of arc BD =?
In the given figure, the OC = 0B as they are radius of triangle, therefore it is a equilateral traiangle.
Hence, the angle COB = 600
`=60xx(π)/(180)` radian
`=(π)/(3)` radian
Now, we know that,
The length of arc, l = radius of the circle, r × angle subtended by the arc
`=20xx(π)/(3)` cm`
`=(20)/(3)xx(22)/(7)` cm
`=20(20)/(21)` cm
This is the length of the arc
Question 6: If in two circles, arcs of the same length subtend angles 600 and 750 at the centre, find the ratio of their radii.
Solution: Let r1 and r2 are the radii of two given circles.
Given, The angle subtend by the arc in one circle = 600
The angle subtend by the arc in second circle = 750
The length of arcs is same.
`θ_1=60°`=`60xx(π)/(180)` radian
`=(π)/(3)` radian
`θ_2=75°`=`75xx(π)/(180)` radian
`=(5π)/(12)` radian
Now, we know that
l = rѲ
In the given circle
`l_1=r_1\θ_1`
`l_2=r_2\θ_2`
As the length of both the arcs are same.
Therefore,
` r_1\θ_1= r_2\θ_2`
Or, `(r_1)/(r_2)=(θ_2)/(θ_1)`
Or, `(r_1)/(r_2)=((5π)/(12))/((π)/(3))`
Or, `(r_1)/(r_2)=5/4`
Therefore, `r_1:r_2=5:4`
Question 7: Find the angle in radian through a pendulum swings if its length is 75cm and the tip describes an arc of length
(i) 10 cm (ii) 15cm (iii) 21cm
Solution:
(i) r = 75cm
length of arc, l = 10cm
The angle Ѳ =?
We know that,
`θ=l/r`
`=(10)/(75)` radian
`=(2)/(15)` radian
(ii) r = 75cm
length of arc, l = 15cm
The angle Ѳ =?
We know that,
`θ=l/r`
`=(15)/(75)` radian
`=(1)/(5)` radian
r = 75cm
length of arc, l = 21cm
The angle Ѳ =?
We know that,
`θ=l/r`
`=(21)/(75)` radian
`=(7)/(25)` radian