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
∴25o= π /180 x 25 radian=5 π/36 radian
(ii) –47° 30'
–47° 30'
=-95/2 degree
Since 180° = π radian
=95/2o = π/180 x (-95/2) radian=(-19/36x2) π radian=-19/72 π radian
∴-47o 30’=-19/72 π radian
(iii) 240°
We know that 180° = π radian
∴520o= 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
=39o+3 x 60/8 minutes [1o=60’]
=39o + 22’ + ½ minutes
=39o22’30” [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=
=-229o+1 x 60/11 minutes [1o=60’]
=-229o+5’+5/11 minutes
=-229o5’27” [1’=60”]
(iii) 5 π/3
We know that π radian = 180°
5 π/3 radian=180/π x 5 π/3 degree=300o
(iv) 7 π /6
We know that π radian = 180°
∴ 7 π /6 radian=180/π x 7 π/6 =210o
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=12o36’ [1o=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 ∆OAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, ∆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 r1 and r2. Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r1, and let an arc of length l subtend an angle of 75° at the centre of the other circle of radius r2.
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=r1π /3 and l=r25π /12
⇒ r1π /3= r25π /12
⇒r1= r25 /4
⇒r1/r2=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