Equal-height towers seen from a point between them: An observer stands between two vertical towers of equal height. The angles of elevation of the nearer and farther tops are 60° and 30° respectively. If the distance to the nearer tower is 100 m, find (i) the height of each tower and (ii) the distance between the towers.

Introduction / Context:A pair of equal-height towers on a straight line with the observer allows immediate height recovery from the nearer 60° sight, then the far distance from the 30° sight, giving the separation by addition.

Given Data / Assumptions:

• Near top elevation = 60° at 100 m.
• Far top elevation = 30°; observer lies between towers.
• Towers have equal height H.

Concept / Approach:Use tan for each: from near tower, H = 100 * tan 60°; from far tower, with distance D_far, H = D_far * tan 30°. Sum near and far horizontal distances to get total gap.

Step-by-Step Solution:

Verification / Alternative check:Check near: H/100 = √3; far: H/300 = 1/√3 → both angles match 60° and 30° respectively.

Why Other Options Are Wrong:Pairs with 100/√3 contradict tan 60°; a total of 300 m separation ignores the far geometry; only “100√3 m and 400 m” is consistent.

Common Pitfalls:Assuming the observer is outside the segment; swapping near/far distances; mixing up tan 30° with cot 30°.

Final Answer:100√3 m and 400 m