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Trigonometry & Heights-Distances

Quantitative Aptitude · Geometry, Mensuration & Trigonometry

Useful for: SSC CGL

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Overview

Standard-angle values, three identities and the tangent ratio for heights solve nearly all trigonometry items.

Key points

Standard values: sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2; tan 30° = 1/√3, tan 45° = 1, tan 60° = √3.
sin²θ + cos²θ = 1; sec²θ − tan²θ = 1; cosec²θ − cot²θ = 1.
sin(90° − θ) = cos θ; tan(90° − θ) = cot θ.
Heights: tan θ = height ÷ horizontal distance.
Angle of elevation from the ground equals the angle of depression from the top (alternate angles).

Explanation

If sin θ = p/h, build the triangle: base = √(h² − p²).
For θ + φ = 90°, sin θ = cos φ and tan θ × tan φ = 1.
Two-angle height problems: write tan for each position and subtract distances.
Maximum of a sin θ + b cos θ is √(a² + b²).

Step by step

Draw the triangle.
Choose the ratio or identity.
Substitute standard values.

Examples

Shadow 20 m, angle of elevation of the Sun 45° → height = 20 × tan 45° = 20 m.
Angle 60°, shadow 10 m → height = 10√3 ≈ 17.32 m.
sin θ = 3/5 → cos θ = 4/5, tan θ = 3/4.

Traps and advanced points

Trap: Using sin instead of tan for height and horizontal distance.
Trap: Mixing up the angle of elevation and the angle at the top.
Trap: Forgetting that sin²θ means (sin θ)².

Quick revision

sin² + cos² = 1.
tan = height/base.
sin(90° − θ) = cos θ.

Practice MCQs

The value of sin 30 degrees + cos 60 degrees is:
1. 1
2. 0
3. 1/2
4. 2
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Answer: (a) 1. sin30 = 1/2, cos60 = 1/2.
If sin θ = 3/5, then cos θ is (θ acute):
1. 5/3
2. 4/5
3. 5/4
4. 3/4
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Answer: (b) 4/5. sin^2+cos^2=1 → cos^2 = 1-9/25 = 16/25 → cos = 4/5.
The value of tan 45 degrees × cos 0 degrees is:
1. 0
2. 2
3. 1/2
4. 1
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Answer: (d) 1. tan45 = 1, cos0 = 1.
If sec θ = 13/12, the value of tan θ is:
1. 5/12
2. 12/5
3. 13/5
4. 5/13
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Answer: (a) 5/12. 1+tan^2 = sec^2 → tan^2 = (13/12)^2 - 1 = 169/144 - 144/144 = 25/144 → tan = 5/12.
The value of (1 - cos^2 θ) / cos^2 θ is:
1. tan^2 θ
2. sin^2 θ
3. sec^2 θ
4. cot^2 θ
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Answer: (a) tan^2 θ. 1-cos^2θ = sin^2θ.
A pole casts a shadow equal to its height. The angle of elevation of the sun is:
1. 45 degrees
2. 30 degrees
3. 60 degrees
4. 90 degrees
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Answer: (a) 45 degrees. tan θ = height/shadow = h/h = 1 → θ = 45 degrees.
The value of cos 90 degrees + sin 90 degrees is:
1. 1/2
2. 1
3. 0
4. 2
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Answer: (b) 1. cos90 = 0, sin90 = 1.
If tan θ = 1, then θ (acute) equals:
1. 60 degrees
2. 90 degrees
3. 45 degrees
4. 30 degrees
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Answer: (c) 45 degrees. tan45 = 1, isliye θ = 45 degrees.
The angle of elevation of the top of a tower from a point 30 m away is 60 degrees. The height of the tower is:
1. 15√3 m
2. 60 m
3. 30√3 m
4. 30 m
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Answer: (c) 30√3 m. tan60 = height/30 → √3 = h/30 → h = 30√3 m.
The value of sin^2 30 + cos^2 30 is:
1. 1
2. 0
3. 1/2
4. √3/2
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Answer: (a) 1. Identity sin^2θ + cos^2θ = 1 for any θ.