Question 1

In a right-angled triangle, the opposite side is 9 cm and the hypotenuse is 15 cm. Find sin θ.

Accepted Answer

Step 1: Recall that sin θ = opposite / hypotenuse. Step 2: Substitute the given values: sin θ = 9 / 15. Step 3: Simplify the fraction: sin θ = 3/5 = 0.6. Therefore, the answer is 3/5 or 0.6.

Question 2

If sin θ = 4/5, what is the value of tan θ (where θ is acute)?

Accepted Answer

Step 1: Use the Pythagorean identity sin²θ + cos²θ = 1. Step 2: Substitute sin θ = 4/5: (4/5)² + cos²θ = 1. Step 3: 16/25 + cos²θ = 1. Step 4: cos²θ = 9/25, so cos θ = 3/5 (positive in acute angle). Step 5: tan θ = sin θ / cos θ = (4/5) / (3/5) = 4/3. Therefore, answer is 4/3.

Question 3

If cot θ = 8/15 and θ is acute, what is the value of cosec²θ - cot²θ?

Accepted Answer

Step 1: We use the identity cosec²θ - cot²θ = 1 (a fundamental trigonometric identity). Step 2: This identity holds for all values of θ where cot θ is defined, regardless of the specific value of cot θ. Step 3: Therefore, cosec²θ - cot²θ = 1.

Question 4

If cot θ = 8/15 and θ lies in the first quadrant, find the value of 17 sin θ - 8 cos θ.

Accepted Answer

Step 1: Given cot θ = 8/15, which means cos θ / sin θ = 8/15, so cos θ = 8k and sin θ = 15k for some positive k. Step 2: Using sin²θ + cos²θ = 1, we get (15k)² + (8k)² = 1 → 225k² + 64k² = 1 → 289k² = 1 → k = 1/17. Step 3: sin θ = 15/17 and cos θ = 8/17. Step 4: Calculate 17 sin θ - 8 cos θ = 17(15/17) - 8(8/17) = 15 - 64/17 = (255 - 64)/17 = 191/17.

Question 5

If sin θ = 3/5 and θ is an acute angle, find the value of tan θ.

Accepted Answer

Step 1: Given sin θ = 3/5. Using sin²θ + cos²θ = 1, we get (3/5)² + cos²θ = 1. Step 2: 9/25 + cos²θ = 1, so cos²θ = 16/25, thus cos θ = 4/5 (positive in acute angle). Step 3: tan θ = sin θ / cos θ = (3/5) ÷ (4/5) = 3/4.

Question 6

In a right-angled triangle, if tan A = 5/12, find the value of sin A × cos A.

Accepted Answer

Step 1: Given tan A = 5/12, which means opposite/adjacent = 5/12. Step 2: Using the Pythagorean theorem, hypotenuse = √(5² + 12²) = √(25 + 144) = √169 = 13. Step 3: sin A = 5/13 and cos A = 12/13. Step 4: sin A × cos A = (5/13) × (12/13) = 60/169.

Question 7

If sin θ + cos θ = √2, find the value of sin³θ + cos³θ.

Accepted Answer

Step 1: Given sin θ + cos θ = √2. Square both sides: sin²θ + 2 sin θ cos θ + cos²θ = 2. Step 2: Since sin²θ + cos²θ = 1, we get 1 + 2 sin θ cos θ = 2, so sin θ cos θ = 1/2. Step 3: Use the identity sin³θ + cos³θ = (sin θ + cos θ)³ - 3 sin θ cos θ(sin θ + cos θ). Step 4: sin³θ + cos³θ = (√2)³ - 3(1/2)(√2) = 2√2 - (3√2)/2 = (4√2 - 3√2)/2 = √2/2.

Question 8

If sec θ - tan θ = 2/5, find the value of sec θ + tan θ.

Accepted Answer

Step 1: Given sec θ - tan θ = 2/5. Use the identity sec²θ - tan²θ = 1, which factors as (sec θ - tan θ)(sec θ + tan θ) = 1. Step 2: Substitute sec θ - tan θ = 2/5 into the identity: (2/5)(sec θ + tan θ) = 1. Step 3: sec θ + tan θ = 5/2.

Question 9

A ladder of length 13 m leans against a wall. If the angle between the ladder and the ground is 67.38°, find the height at which the ladder touches the wall. (Given: sin 67.38° ≈ 0.92, cos 67.38° ≈ 0.39)

Accepted Answer

Step 1: The ladder forms the hypotenuse of a right triangle. The angle between the ladder and the ground is 67.38°. Step 2: The height at which the ladder touches the wall is the side opposite to this angle. Step 3: Using sin θ = opposite/hypotenuse, we have sin 67.38° = height/13. Step 4: height = 13 × sin 67.38° = 13 × 0.92 = 11.96 m ≈ 12 m.

SSC CHSL Basic Trig Ratios

Basic Trig Ratios— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Basic Trig Ratios — Practice Questions

60-Second Revision — Basic Trig Ratios