Question 1

Q1: Find the radian measures corresponding to the following degree measures: (i) 25° (ii) -47° 30' (iii) 240° (iv) 520°

Accepted Answer

We know that 180° = π radian. Therefore, to convert degrees to radians, we multiply the degree measure by π/180. (i) 25° Radian measure = 25 × (π/180) = 5π/36 (ii) -47° 30' First, we convert 30' to degrees. Since 60' = 1°, 30' = (30/60)° = 0.5°. So, -47° 30' = -(47 + 0.5)° = -47.5° = -95/2 degrees....

Question 2

Q2: 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

Accepted Answer

We know that π radian = 180°. Therefore, to convert radians to degrees, we multiply the radian measure by 180/π. (i) 11/16 Degree measure = (11/16) × (180/π) = (11/16) × (180 × 7 / 22) = (1/16) × (180 × 7 / 2) = (90 × 7) / 16 = (45 × 7) / 8 = 315/8 degrees. 315/8° = 39 3/8° = 39° + (3/8) × 60' = 39°...

Question 3

Q3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Accepted Answer

The number of revolutions made by the wheel in 1 minute (60 seconds) is 360. Number of revolutions in 1 second = 360 / 60 = 6 revolutions. We know that in one complete revolution, the angle turned is 2π radians. Therefore, the angle turned in 6 revolutions = 6 × 2π radians. Angle turned in 1 second...

Question 4

Q4: 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).

Accepted Answer

Given: Radius of the circle, r = 100 cm Length of the arc, l = 22 cm We know the relationship between the angle (θ in radians), arc length (l), and radius (r) is θ = l/r. θ = 22 / 100 radians. To convert this angle to degrees, we multiply by 180/π. Degree measure = (22/100) × (180/π) Using π = 22/7:...

Question 5

Q5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Accepted Answer

Given: Diameter of the circle = 40 cm Radius of the circle, r = Diameter / 2 = 40 / 2 = 20 cm. Length of the chord = 20 cm. Consider the triangle formed by the chord and the two radii connecting the endpoints of the chord to the center of the circle. Let the center be O and the endpoints of the chor...

Question 6

Q6: If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Accepted Answer

Let the two circles have radii r₁ and r₂, respectively. Let the length of the arc be 'l' for both circles. For the first circle: Angle, θ₁ = 60° Radius = r₁ For the second circle: Angle, θ₂ = 75° Radius = r₂ We know the formula l = rθ, where θ must be in radians. First, convert the angles to radians...

Question 7

Q7: Find the angle in radian through 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 c...

Accepted Answer

The length of the pendulum acts as the radius (r) of the circle, and the distance the tip moves is the arc length (l). Given, r = 75 cm. The angle in radians (θ) is given by the formula θ = l/r. (i) l = 10 cm θ = 10 / 75 = 2/15 radians (ii) l = 15 cm θ = 15 / 75 = 1/5 radians (iii) l = 21 cm θ = 21...

Question 8

Q1: Find the values of other five trigonometric functions in Exercises 1 to 5. 1. cos x = -1/2, x lies in third quadrant.

Accepted Answer

Given: cos x = -1/2 and x lies in the third quadrant. In the third quadrant, tan x and cot x are positive, while sin x, cos x, sec x, and cosec x are negative. 1.  sec x:     sec x = 1 / cos x = 1 / (-1/2) = -2. 2.  sin x:     We know that sin²x + cos²x = 1.     sin²x = 1 - cos²x = 1 - (-1/2)² = 1 -...

Question 9

Q10: cot (-15π/4)

Accepted Answer

We know that cot(-x) = -cot(x). Also, the values of cot x repeat after an interval of π radians. cot(-15π/4) = -cot(15π/4) Now, we write 15π/4 as a multiple of π. 15π/4 = (16π - π)/4 = 16π/4 - π/4 = 4π - π/4 -cot(15π/4) = -cot(4π - π/4) Since cot(nπ + x) = cot(x): -cot(4π - π/4) = -cot(-π/4) Using c...

Question 10

Q2: sin x = 3/5, x lies in second quadrant.

Accepted Answer

Given: sin x = 3/5 and x lies in the second quadrant. In the second quadrant, sin x and cosec x are positive, while cos x, tan x, sec x, and cot x are negative. 1.  cosec x:     cosec x = 1 / sin x = 1 / (3/5) = 5/3. 2.  cos x:     We know that sin²x + cos²x = 1.     cos²x = 1 - sin²x = 1 - (3/5)² =...

Question 11

Q3: cot x = 3/4, x lies in third quadrant.

Accepted Answer

Given: cot x = 3/4 and x lies in the third quadrant. In the third quadrant, tan x and cot x are positive, while sin x, cos x, sec x, and cosec x are negative. 1.  tan x:     tan x = 1 / cot x = 1 / (3/4) = 4/3. 2.  cosec x:     We know that 1 + cot²x = cosec²x.     cosec²x = 1 + (3/4)² = 1 + 9/16 =...

Question 12

Q4: sec x = 13/5, x lies in fourth quadrant.

Accepted Answer

Given: sec x = 13/5 and x lies in the fourth quadrant. In the fourth quadrant, cos x and sec x are positive, while sin x, tan x, cosec x, and cot x are negative. 1.  cos x:     cos x = 1 / sec x = 1 / (13/5) = 5/13. 2.  sin x:     We know that sin²x + cos²x = 1.     sin²x = 1 - cos²x = 1 - (5/13)² =...

Question 13

Q5: tan x = -5/12, x lies in second quadrant.

Accepted Answer

Given: tan x = -5/12 and x lies in the second quadrant. In the second quadrant, sin x and cosec x are positive, while cos x, tan x, sec x, and cot x are negative. 1.  cot x:     cot x = 1 / tan x = 1 / (-5/12) = -12/5. 2.  sec x:     We know that 1 + tan²x = sec²x.     sec²x = 1 + (-5/12)² = 1 + 25/...

Question 14

Q6: Find the values of the trigonometric functions in Exercises 6 to 10. 6. sin 765°

Accepted Answer

We know that the values of sin x repeat after an interval of 360° or 2π radians. We can write 765° as a multiple of 360° plus a remainder. 765° = 2 × 360° + 45° = 720° + 45° Therefore, sin(765°) = sin(2 × 360° + 45°) = sin(45°). Since sin(45°) = 1/√2, sin(765°) = 1/√2.

Question 15

Q7: cosec(-1410°)

Accepted Answer

We know that cosec(-x) = -cosec(x). Also, the values of cosec x repeat after an interval of 360°. cosec(-1410°) = -cosec(1410°) Now, we write 1410° as a multiple of 360° plus a remainder. 1410° = 3 × 360° + 330° = 1080° + 330° Alternatively, 1410° = 4 × 360° - 30° = 1440° - 30° Using the second form...

Question 16

Q8: tan (19π/3)

Accepted Answer

We know that the values of tan x repeat after an interval of π radians. We can write 19π/3 as a multiple of π plus a remainder. 19π/3 = (18π + π)/3 = 18π/3 + π/3 = 6π + π/3 Therefore, tan(19π/3) = tan(6π + π/3). Since tan(nπ + x) = tan(x) for any integer n: tan(6π + π/3) = tan(π/3). We know that tan...

Question 17

Q9: sin (-11π/3)

Accepted Answer

We know that sin(-x) = -sin(x). Also, the values of sin x repeat after an interval of 2π radians. sin(-11π/3) = -sin(11π/3) Now, we write 11π/3 as a multiple of 2π. 11π/3 = (12π - π)/3 = 12π/3 - π/3 = 4π - π/3 = 2 × 2π - π/3 -sin(11π/3) = -sin(4π - π/3) Since sin(2nπ + x) = sin(x): -sin(4π - π/3) =...

Question 18

Q1: Prove that:  1. sin²(π/6) + cos²(π/3) - tan²(π/4) = -1/2

Accepted Answer

To prove the identity, we will evaluate the Left Hand Side (L.H.S.) by substituting the known values of the trigonometric functions. L.H.S. = sin²(π/6) + cos²(π/3) - tan²(π/4) We know the values:    sin(π/6) = 1/2    cos(π/3) = 1/2 *   tan(π/4) = 1 Substitute these values into the L.H.S.: L.H.S. = (...

Question 19

Q10: 10. sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x = cos x

Accepted Answer

We will use the trigonometric identity: cos A cos B + sin A sin B = cos(A - B) The Left Hand Side (L.H.S.) of the given equation is in the form of this identity. L.H.S. = sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x Let A = (n+2)x and B = (n+1)x. Rewriting the L.H.S. to match the identity form: L.H.S....

Question 20

Q11: 11. cos(3π/4 + x) - cos(3π/4 - x) = -√2 sin x

Accepted Answer

We will use the sum-to-product identity: cos A - cos B = -2 sin((A + B)/2) sin((A - B)/2) Let A = (3π/4 + x) and B = (3π/4 - x). The Left Hand Side (L.H.S.) of the given equation is in the form of this identity. L.H.S. = cos(3π/4 + x) - cos(3π/4 - x) First, we find (A + B)/2 and (A - B)/2:    A + B...

NCERT Solutions