NCERT Solutions Class 11 Mathematics Trigonometric-Functions Exercise 3.1

Question 1:
Find the radian measures corresponding to the following degree measures:
(i) 25°                            (ii) – 47° 30′                           (iii) 240°                      (iv) 520°

Answer:
(i) 25°
We know that 180° = π radian
So, 25° = (π/180) * 25 radian = 25π/180 radian = 5π/36 radian
(ii) -47° 30′
-47° 30′ = -47½ = -95/2 degree
Since 180° = π radian
So, -95/2 degree = (π/180) * (-95/2) radian
                               = -19π/(36 * 2)
                               = -19π/72 radian
So, -47° 30′ = -19π/72 radian
(iii) 240°
We know that 180° = π radian
So, 240° = (π/180) * 240 radian = 4π/3 radian
(iv) 520°
We know that 180° = π radian
So, 520° = (π/180) * 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:
(i) 11/16
We know that π radian = 180°
So, 11/16 radian = (180/π) * (11/16) degree
                               = (45 * 11)/(π * 4) degree
                               = (45 * 11 * 7)/(22 * 4) degree
                               = 315/8 degree
                               = 39 ³/₈ degree
                               = 39° + (3 * 60)/8 minutes             [Since 1° = 60’]      
                               = 39° + 22’ + 1/2 minutes
                               = 39° 22’ 30’’                              [Since 1’ = 60’’] 
(ii) -4
We know that π radian = 180°
So, -4 radian = (180/π) * (-4) degree
                        = {180 * (-4) * 7}/22 degree
                         = -2520/11 degree
                         = -229 ¹/₁₁ degree
                         = -229° + (1 * 60)/11 minutes             [Since 1° = 60’]      
                         = -229° + 5’ + 5/11 minutes
                         = -229° 5’ 27’’                                         [Since 1’ = 60’’] 
 
(iii) 5π/3
We know that π radian = 180°
So, 5π/3 = (180/ π) * (5π/3) = 300°
 
(iv) 7π/6
We know that π radian = 180°
So, 7π/6 = (180/ π) * (7π/6) = 210°

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

Answer:
Number of revolutions made by the wheel in 1 minute = 360
So, the number of revolutions made by the wheel in 1 second = 360/60 = 6
In one complete revolution, the wheel turns an angle of 2π radian.
Hence, in 6 complete revolutions, it will turn 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:
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
Therefore, for r = 100 cm, l = 22 cm, we have
θ = 22/100 radian = (180/π) * (22/100) degree
                                 = (180 * 22 * 7)/(22 * 100) degree
                                 = 126/10 degree
                                 = 12 ³/₅ degree
                                 = 12° 36’                      [Since 1° = 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:
Diameter of the circle = 40 cm
Radius (r) of the circle = 40/2 = 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
⇒ π/3 = AB/20
⇒ AB = 20π/3
Thus, the length of the minor arc of the chord is 20π/3 cm.