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ABC is a triangular park with AB = AC = 100 m. A television tower stands at the midpoint of BC. The angles of elevation of the top of the tower at A, B, C are 45°, 60°, 60°, respectively. Find the height of the tower.

ABC is a triangular park with AB = AC = 100 m. A television tower stands at the midpoint of BC. The angles of elevation of the top of the tower at A, B, C are 45°, 60°, 60°, respectively. Find the height of the tower.

Grade:10

1 Answers

Jitender Pal
askIITians Faculty 365 Points
8 years ago
Hello Student,
Please find the answer to your question
Let ABC be the triangle region with AB = AC = 100m Let M be the mid point of BC at which tower LM stands.
As ∆ABC is isosceles and M is mid pt. of BC
∴ AM ⊥ BC.
Let LM = h be the ht. of tower.
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In ∆ALM, tan 45° = LM/MA ⇒ LM = MA
∴ MA = h
Also in ∆BLM, tan 60° = LM/BM
⇒ √3 = h/BM ⇒ BM = h/√3
Now in rt ∆AMB, we have, AB2 = AM2 + BM2
⇒ (100)2 = h2 + h2/3
⇒ 4h2/3 = 10000
⇒ h = 50 √3 m.

Thanks
Jitender Pal
askIITians Faculty

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