Isosceles park, tower at midpoint — mixed inverse trig data In triangle ABC with AB = AC = 100 m, point M is the midpoint of BC, where a clock tower stands. The angle of elevation of the tower top from A is cot⁻¹(3.2), and from B is cosec⁻¹(2.6). Find the height of the tower (in meters).

Introduction / Context:This problem combines geometry of an isosceles triangle with trigonometric angles given in inverse functions. Placing a tower at the midpoint of the base leverages a right triangle from each vertex to the midpoint.

Given Data / Assumptions:

• AB = AC = 100 m; M is midpoint of BC.
• Let tower height be H at M.
• At A: angle of elevation α = cot⁻¹(3.2) ⇒ tan α = 1/3.2 = 0.3125.
• At B: angle of elevation β = cosec⁻¹(2.6) ⇒ sin β = 1/2.6 ⇒ tan β = sin/√(1 − sin²) = (1/2.6)/√(1 − 1/2.6²) = 5/12.

Concept / Approach:Let AM be the altitude from A to BC, and BM = CM be half the base. In right triangle ABM: AM and BM are perpendicular. From A: H/AM = tan α = 5/16 ⇒ H = (5/16)AM. From B: H/BM = tan β = 5/12 ⇒ H = (5/12)BM. Use AB² = AM² + BM² = 100² to eliminate AM and BM.

Step-by-Step Solution:

Verification / Alternative check:Back-substitution yields consistent α and β values.

Why Other Options Are Wrong:12.5 m and 50 m do not satisfy both elevation relationships and the 100 m side length simultaneously.

Common Pitfalls:Confusing “angle subtended” with “angle of elevation,” and mishandling inverse trig conversions.

Final Answer:25