A vertical pole TF has its base F on the ground. From a point A, 25 meters from F, the angle of elevation of T is x°. From another point B, 75 meters from F (and in the same line as A), the angle of elevation of T is y°. If the height of the pole is 25√3 meters, what is the value of x° + y°?

Correct — C. Use tan(angle of elevation) = height ÷ distance from the base F.

At A: tan x° = 25√3 ÷ 25 = √3, so x = 60°
At B: tan y° = 25√3 ÷ 75 = √3⁄3 = 1⁄√3, so y = 30°

Add the two angles:
x + y = 60° + 30° = 90° → option (c)

Why the others are wrong

• (a)
  60° — 60° is x alone, the angle at A. The question asks for x + y, and B adds its own 30°.
• (b)
  75° — With y = 30°, a sum of 75° needs x = 45°, which happens only when distance equals height. But 25√3 ≈ 43.3 m is not 25 m, so tan x = √3 and x = 60°.
• (d)
  120° — 120° is 60° + 60°, as if B saw the top at the same angle as A. B is three times as far from F, so its angle must be smaller: tan y = 1⁄√3 gives 30°.

For two points at distances a and b from the base of a pole of height h, tan x = h⁄a and tan y = h⁄b.

Multiply them: tan x × tan y = h²⁄(ab). If h² = ab, the product is 1, and two acute angles whose tangents multiply to 1 are complementary, so x + y = 90°.

Here h² = (25√3)² = 1875 and ab = 25 × 75 = 1875, so the sum is 90° even before either angle is found.

Key facts

• tan 60° = √3 and tan 30° = 1⁄√3.
• If tan x × tan y = 1 for acute angles x and y, then x + y = 90°.
• From distances a and b on a line through the base, the two elevations are complementary exactly when height² = a × b.

Common traps

• Stopping at x = 60° and marking the first option that matches it.
• Reading 25√3 as 25, which makes tan x = 1 and x = 45°.

No directly related past PYQ was found.