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HEIGHTS AND DISTANCES — COMPLETE GUIDE FOR SSC CGL ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
CORE CONCEPT ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Heights and Distances is a real-life application of Trigonometry. We use it to find the height of a building, tower, tree, or mountain — without physically measuring it. We also find the distance between two objects. The tool we use is the angle formed between the line of sight and the horizontal ground. Think of it this way: You stand on the ground and look up at the top of a tower. The angle your eyes make with the horizontal is the Angle of Elevation. Now imagine you are standing on the top of a building and looking DOWN at a car on the road. That angle is the Angle of Depression.
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ KEY DEFINITIONS
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ • Angle of Elevation: The angle formed when you look UPWARD from the horizontal. It is measured from the horizontal line UP to the line of sight.
• Angle of Depression: The angle formed when you look DOWNWARD from the horizontal. It is measured from the horizontal line DOWN to the line of sight. • Line of Sight: The straight line from your eye to the object.
• Horizontal Line: The flat line parallel to the ground from your eye level.
The Angle of Elevation from point A to point B equals the Angle of Depression from point B to point A. They are alternate interior angles and are always EQUAL. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
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Key Trigonometric Values (MEMORISE THESE):
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— WHAT SSC ASKS ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SSC CGL asks 1–2 questions from this topic every year. The most common patterns are: 1. One observer, one tower — find height or distance. 2. Two angles of elevation from two different points — find height. 3.
Tower on top of a hill or building — combined height problems. 4. Two observers on opposite sides of a tower. 5. Shadow-based problems using sun's angle of elevation. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SHORTCUTS AND TRICKS ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SHORTCUT 1 — Two Angle Formula (Most Tested): When a tower of height H is observed from two points at distances d1 and d2 on the same side, with angles A and B: H² = d1 × d2 × tan(A) × tan(B) Special case: If angles are complementary (A + B = 90°), then H = √(d1 × d2) SHORTCUT 2 — 45° Shortcut: When angle of elevation = 45°, tan 45° = 1, so Height = Distance. This is the fastest case.
No calculation needed. SHORTCUT 3 — Shadow Length Trick: When the sun makes angle θ with the ground: Height of object = Shadow length × tan(θ) If θ = 60°, Height = Shadow × √3 If θ = 30°, Height = Shadow / √3 ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Draw the situation. You have a right triangle. The base (horizontal distance) = 40 m. The angle at the base = 30°. Height = H (unknown).
Use formula — tan(angle) = Height / Distance tan 30° = H / 40 1/√3 = H / 40 H = 40 / √3 H = 40√3 / 3 H = 40 × 1.732 / 3 ≈ 23.09 m Final Answer: Height of tower = 40/√3 = 40√3/3 meters ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ WORKED EXAMPLE 2 — Two Angle Problem ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Question: The angle of elevation of the top of a tower from point A is 45°. From point B, which is 20 m closer to the tower, the angle of elevation is 60°. Find the height of the tower.
Let height = H. Let distance from B to base of tower = d. From B: tan 60° = H/d → √3 = H/d → d = H/√3 From A: tan 45° = H/(d + 20) → 1 = H/(d + 20) → d + 20 = H
Substitute d = H/√3 into second equation. H/√3 + 20 = H 20 = H − H/√3 20 = H(1 − 1/√3) 20 = H × (√3 − 1)/√3 H = 20√3 / (√3 − 1)
Rationalise — Multiply numerator and denominator by (√3 + 1). H = 20√3(√3 + 1) / [(√3 − 1)(√3 + 1)] H = 20√3(√3 + 1) / (3 − 1) H = 20√3(√3 + 1) / 2 H = 10√3(√3 + 1) H = 10(3 + √3) H = 30 + 10√3 meters Final Answer: Height = 10(3 + √3) = approximately 47.32 m ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
— THE NUMBER 1 TRAP ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Students confuse Angle of Depression with Angle of Elevation in two-observer problems. They draw the angle of depression from the wrong position or forget that the angle of depression from the top equals the angle of elevation from the bottom. Always draw the figure first.
Never solve without a diagram. Many errors are simply diagram errors, not calculation errors.
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