Exercise 3.1 answers: Trigonometric Functions
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Exercise 3.1
7 questions · page 48 of the book
Question 1
“Find the radian measures corresponding to the following degree measures” · p. 48
Open NCERT p. 48Matches NCERT’s answer
(i) 25°
- 25° in radians = 25 × π/180.
- Cancel the common factor 5: = 5π/36.
Answer5π/36 rad
(ii) –47°30′
- 30′ = ½°, so –47°30′ = –47.5° = –95/2°.
- Multiply by π/180: (–95/2) × π/180 = –19π/72.
Answer–19π/72 rad
(iii) 240°
- 240° in radians = 240 × π/180.
- Simplify: = 4π/3.
Answer4π/3 rad
(iv) 520°
- 520° in radians = 520 × π/180.
- Simplify: = 26π/9.
Answer26π/9 rad
Watch this explained “The pile that only pays”, 12:27 into The exchange rate between the two units, and the arc-length rule it buys
Question 2
“Find the degree measures corresponding to the following radian measures” · p. 49
Open NCERT p. 49Matches NCERT’s answer
(i) 11/16
- Degrees = radians × 180/π.
- With π = 22/7, 180/π = 1260/22 = 630/11.
- 11/16 × 630/11 = 630/16 = 315/8.
Answer315/8° (= 39°22′30″)
(ii) –4
- Degrees = –4 × 630/11 = –2520/11.
Answer–2520/11° (≈ –229°5′27″)
(iii) 5π/3
- 5π/3 is already a multiple of π, so π cancels: 5π/3 × 180/π = 5 × 60 = 300.
Answer300°
(iv) 7π/6
- 7π/6 × 180/π = 7 × 30 = 210.
Answer210°
Watch this explained “The pile that only pays”, 12:27 into The exchange rate between the two units, and the arc-length rule it buys
Question 3
“A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?” · p. 49
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- One revolution is a full turn = 2π radians.
- 360 revolutions in 60 seconds = 360 × 2π radians in 60 seconds.
- In one second: (720π)/60 = 12π radians.
Answer12π radians
Watch this explained “The pile that only pays”, 12:27 into The exchange rate between the two units, and the arc-length rule it buys
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” · p. 49
Open NCERT p. 49Matches NCERT’s answer
- Angle in radians = arc length ÷ radius = 22/100.
- Convert to degrees: (22/100) × 180/π, using π = 22/7.
- = (22/100) × (180 × 7/22) = 1260/100 = 12.6° = 63/5°.
Answer63/5° = 12°36′
Watch this explained “The pile that collects”, 14:01 into The exchange rate between the two units, and the arc-length rule it buys
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.” · p. 49
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- Radius = diameter ÷ 2 = 20 cm, which is the same as the chord length (20 cm).
- So the triangle made by the two radii and the chord has all three sides equal — it is equilateral.
- Every angle of an equilateral triangle is 60°, so the angle at the centre is 60° = π/3 radians.
- Arc length = radius × angle (in radians) = 20 × π/3.
Answer20π/3 cm
Watch this explained “The pile that collects”, 14:01 into The exchange rate between the two units, and the arc-length rule it buys
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.” · p. 49
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- Let the radii be r₁ and r₂, and let l be the common arc length.
- l = r₁θ₁ = r₂θ₂, so r₁/r₂ = θ₂/θ₁.
- The unit of the angles cancels in the ratio, so r₁/r₂ = 75/60 = 5/4.
Answerr₁ : r₂ = 5 : 4
Watch this explained “One arc, two circles”, 11:11 into The exchange rate between the two units, and the arc-length rule it buys
Question 7
“Find the angle in radian through which a pendulum swings if its length is 75 cm and … describes an arc of length” · p. 49
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(i) 10 cm
- θ = arc ÷ radius = 10/75 = 2/15.
Answer2/15 rad
(ii) 15 cm
- θ = 15/75 = 1/5.
Answer1/5 rad
(iii) 21 cm
- θ = 21/75 = 7/25.
Answer7/25 rad
Watch this explained “The pile that collects”, 14:01 into The exchange rate between the two units, and the arc-length rule it buys
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