Trigonometry
Class 10 Maths · Trigonometry · 29 questions with answers
Trigonometry is the chapter that most rewards being systematic. The standard values are a small table, the identities all come from three relationships, and heights and distances always start with a labelled diagram. Almost nothing here needs insight - it needs the table known cold and the diagram drawn first.
- Recalls the standard values without hesitation
- Evaluates an expression in standard angles
- Proves an identity by converting everything to sine and cosine
- Uses complementary angles to simplify
- Solves a height and distance problem from a labelled diagram
Exercise 1 of 3 · 10 questions
Evaluate.
Write the answer · Warm-up
Here’s one done for you
In the last one, note that sec 30° = 2/√3, so sec² 30° = 4/3 rather than 2/√3. Squaring the whole ratio - not just part of it - is where this question is usually lost. Writing sec 30° as its value on a separate line before squaring takes two seconds and prevents it.
- 1)sin 30° cos 60° + cos 30° sin 60°
- 2)2 tan² 45° + cos² 30° - sin² 60°
- 3)(sin 30° + cos 60°) ÷ tan 45°
- 4)cos² 45° + sin² 45°
- 5)(5 cos² 60° + 4 sec² 30° - tan² 45°) ÷ (sin² 30° + cos² 30°)
- 6)2 sin 30° + 3 cos 60°
- 7)sin 60° cos 30° - cos 60° sin 30°
- 8)tan 45° + cot 45°
- 9)sin² 30° + cos² 30°
- 10)(tan 60° − tan 30°) ÷ (1 + tan 60° tan 30°)
Answers
- 1) (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
- 2) 2(1) + 3/4 - 3/4 = 2.
- 3) (1/2 + 1/2) ÷ 1 = 1.
- 4) 1/2 + 1/2 = 1, which is the identity sin²θ + cos²θ = 1 at 45°.
- 5) The denominator is 1. The numerator is 5(1/4) + 4(4/3) - 1 = 5/4 + 16/3 - 1 = 67/12.
- 6) 2(1/2) + 3(1/2) = 1 + 1.5 = 2.5.
- 7) (√3/2)(√3/2) - (1/2)(1/2) = 3/4 - 1/4 = 1/2. It is sin(60° - 30°) = sin 30°.
- 8) 1 + 1 = 2.
- 9) 1, since sin²θ + cos²θ = 1 for any angle.
- 10) (√3 - 1/√3) ÷ (1 + 1) = (2/√3) ÷ 2 = 1/√3. It is tan(60° - 30°) = tan 30°.
Exercise 2 of 3 · 8 questions
Prove each identity. Work on one side only.
Write the answer · Guided
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The fourth proof is the model for the whole chapter. Convert everything to sine and cosine, then simplify - (1 - sinθ)(1 + sinθ) becomes 1 - sin²θ, which is cos²θ, which cancels. When an identity looks impossible, converting to sine and cosine is almost always the move, and 'almost always' is close enough to always in a board paper.
- 1)(1 - cos²θ) cosec²θ = 1
- 2)(1 + tan²A) ÷ (1 + cot²A) = tan²A
- 3)(sin A + cos A)² + (sin A - cos A)² = 2
- 4)sec θ (1 - sin θ)(sec θ + tan θ) = 1
- 5)Prove (1 + cot²A) sin²A = 1.
- 6)Prove (sec A - tan A)(sec A + tan A) = 1.
- 7)Prove (1 - sin²θ) sec²θ = 1.
- 8)Prove tan θ + cot θ = sec θ cosec θ.
Answers
- 1) LHS = sin²θ × (1/sin²θ) = 1 = RHS, using 1 - cos²θ = sin²θ.
- 2) LHS = sec²A ÷ cosec²A = (1/cos²A) ÷ (1/sin²A) = sin²A/cos²A = tan²A = RHS.
- 3) Expanding gives sin²A + 2 sinA cosA + cos²A + sin²A - 2 sinA cosA + cos²A = 2(sin²A + cos²A) = 2.
- 4) LHS = (1/cosθ)(1 - sinθ)(1/cosθ + sinθ/cosθ) = (1 - sinθ)(1 + sinθ)/cos²θ = (1 - sin²θ)/cos²θ = cos²θ/cos²θ = 1.
- 5) 1 + cot²A = cosec²A, and cosec²A × sin²A = (1/sin²A) × sin²A = 1.
- 6) The left side is sec²A - tan²A, and sec²A - tan²A = 1 is a standard identity.
- 7) 1 - sin²θ = cos²θ, and cos²θ × sec²θ = cos²θ × (1/cos²θ) = 1.
- 8) tan θ + cot θ = sin/cos + cos/sin = (sin² + cos²)/(sin cos) = 1/(sin cos) = sec θ cosec θ.
Exercise 3 of 3 · 11 questions
Draw the diagram first, then solve. Use complementary angles where they help.
Write the answer · Practice
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For the kite, the height is opposite the angle and the string is the hypotenuse, so sine is the ratio to use - 60 = L sin 60°. Choosing the wrong ratio is the main way these questions go wrong, and a labelled diagram prevents it: mark which side is opposite, which is adjacent and which is the hypotenuse before writing anything. Thirty seconds of diagram saves the whole question.
- 1)Evaluate sin 18° ÷ cos 72°.
- 2)Evaluate tan 26° ÷ cot 64°.
- 3)Evaluate sin 25° cos 65° + cos 25° sin 65°.
- 4)The shadow of a tower is √3 times its height. Find the sun's angle of elevation.
- 5)A ladder 10 m long leans against a wall at 60° to the ground. How high up the wall does it reach?
- 6)A kite is flying at a height of 60 m with the string making 60° with the horizontal. Find the length of the string.
- 7)Evaluate cos 40° ÷ sin 50°.
- 8)Evaluate tan 15° tan 75°.
- 9)Evaluate sec 70° ÷ cosec 20°.
- 10)A pole 12 m high casts a shadow 12 m long. Find the sun's angle of elevation.
- 11)From the top of a 30 m building the angle of depression of a car is 30°. Find the distance of the car from the foot of the building.
Answers
- 1) 1. Since cos 72° = cos(90° - 18°) = sin 18°.
- 2) 1. Since cot 64° = cot(90° - 26°) = tan 26°.
- 3) 1. Since cos 65° = sin 25° and sin 65° = cos 25°, the expression becomes sin² 25° + cos² 25° = 1.
- 4) 30°. If the height is h, the shadow is √3h, so tan θ = h/(√3h) = 1/√3, giving θ = 30°.
- 5) 5√3 m, about 8.66 m. The height is 10 sin 60° = 10 × √3/2.
- 6) 40√3 m, about 69.3 m. The height is opposite the 60° angle, so 60 = L sin 60°, giving L = 60 ÷ (√3/2) = 120/√3.
- 7) 1. cos 40° = sin 50°, because they are complementary.
- 8) 1. tan 75° = cot 15°, and tan 15° cot 15° = 1.
- 9) 1. sec 70° = cosec 20°, because 70° and 20° are complementary.
- 10) tan θ = 12/12 = 1, so θ = 45°.
- 11) tan 30° = 30/d, so d = 30 ÷ (1/√3) = 30√3 m, about 51.96 m.
While your child works
- The standard values table must be automatic. Everything else in the chapter depends on it, and hesitation there slows every question.
- When an identity looks impossible, convert everything to sine and cosine. It works nearly every time.
- For heights and distances, draw and label the diagram before writing any equation. Choosing the wrong ratio is the main error.