A vertical pole of height H stands on the ground. From a point P on the ground, the angle of elevation of the top of the pole is 60°. From a different point Q, located 10 meters away from point P (directly along the line extending from the base of the pole), the angle of elevation to the top of the pole is 30°.What is the height of the pole?

Correct — A.
The angle falls from 60° at P to 30° at Q, so Q is 10 m farther out on the same line.
Base to P: H ÷ tan 60° = H⁄√3
Base to Q: H ÷ tan 30° = H√3

Gap PQ: H√3 − H⁄√3 = 3H⁄√3 − H⁄√3 = 2H⁄√3
2H⁄√3 = 10, so H = 10√3 ÷ 2 = 5√3 m → option (a)

Why the others are wrong

• (b)
  10√3m — 10√3 m would put P 10 m and Q 30 m from the base, a gap of 20 m. The stem fixes the gap at 10 m, which halves the height.
• (c)
  10 m — 10 m is the gap PQ, and also the sight line from P to the top, not the height. A 10 m pole would make the gap 2 × 10⁄√3 ≈ 11.5 m.
• (d)
  20 m — 20 m would make the gap between P and Q 2 × 20⁄√3 ≈ 23.1 m, more than double the 10 m stated.

Two angles of elevation from two points on one line give two right triangles that share the height H. Each base distance is H ÷ tan(angle).

The smaller angle belongs to the farther point, so Q lies beyond P and the 10 m is the difference of the two base distances, not their sum.

Key facts

• tan 60° = √3 and tan 30° = 1⁄√3.
• With elevations of 60° and 30°, the base distances are H⁄√3 and H√3, in the ratio 1 : 3.
• Here P is 5 m and Q is 15 m from the base, and H = 5√3 ≈ 8.66 m.

Common traps

• Using sin or cos where tan is needed: the height and the ground distance are the two legs
• Adding the two base distances when both points are on the same side of the pole
• Stopping at 10 m, the sight line from P, instead of taking its vertical part

No directly related past PYQ was found.