ABC is a triangular park with AB = AC = 100 m . A TV tower stands at the mid-point of BC . The angles of elevation of the top of the tower at A, B, C are 4 5 ∘ , 6 0 ∘ , 6 0 ∘ respectively. The height

Question

ABC is a triangular park with AB = AC = 100 m . A TV tower stands at the mid-point of BC . The angles of elevation of the top of the tower at A, B, C are 4 5 ∘ , 6 0 ∘ , 6 0 ∘ respectively. The height of the tower is

TestHub · Accepted Answer

tan 4 5 ∘ = AQ PQ ​ = AQ h ​ ⇒ h = AQ Where PQ is tower and ABC is the park, with Q being mid point of the side BC and PQ = h Also, A Q 2 + B Q 2 = 10 0 2 ⇒ h 2 + h 2 cot 2 6 0 ∘ = 10 0 2 ⇒ h 2 [ 1 + 3 1 ​ ] = 10 0 2 ⇒ h 2 = 4 3 × 10 0 2 ​ ⇒ h = 50 3 ​

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