A tower is 20 meters tall. What is the length of the shadow cast by the tower when the sun's elevation is 45°?

Correct — B.
Set up the right triangle: tower = 20 m (opposite), shadow = s (adjacent)

Use tangent: tan 45° = 20 ⁄ s
Substitute tan 45° = 1: 1 = 20 ⁄ s
Solve for s: s = 20 m → option (b)

Why the others are wrong

• (a)
  10m — 10m makes the shadow half the height, which needs tan θ = 20 ⁄ 10 = 2, a sun well above 45°. At 45°, tan θ = 1.
• (c)
  20√3 m — 20√3 m is the shadow at a 30° elevation, where tan 30° = 1⁄√3 and s = 20 ÷ (1⁄√3). At 45° the shadow shrinks to 20 m.
• (d)
  10√3 m — 10√3 m ≈ 17.3 m would need tan θ = 20 ⁄ (10√3) = 2⁄√3 ≈ 1.15. At 45°, tan θ = 1, so the shadow equals the height.

Heights and distances reduce to one right triangle. The object is the opposite side, the shadow is the adjacent side, and the angle of elevation sits at the tip of the shadow.

So tan θ = height ⁄ shadow. At 45°, tan θ = 1, the triangle is isosceles, and the shadow equals the height. At 30° the shadow is height × √3; at 60° it is height ⁄ √3.

Key facts

• tan θ = opposite ⁄ adjacent = height ⁄ shadow
• tan 30° = 1⁄√3, tan 45° = 1, tan 60° = √3
• At a 45° elevation the shadow equals the height

Common traps

• Using sin or cos where height and shadow call for tan
• Inverting the ratio and writing tan θ = shadow ⁄ height

No directly related past PYQ was found.