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An observer at O notices that the angle of elevation of the top of a tower is 30°. The line joining O to the base of the tower makes an angle of tan -1 (1/√2) with the North and is inclined Eastwards. The observer travels a distance of 300 meters towards the North to a point A and finds the tower to his East. The angle of elevation of the top of the tower at A is ϕ, Find ϕ and the height of the tower

An observer at O notices that the angle of elevation of the top of a tower is 30°. The line joining O to the base of the tower makes an angle of tan-1 (1/√2) with the North and is inclined Eastwards. The observer travels a distance of 300 meters towards the North to a point A and finds the tower to his East. The angle of elevation of the top of the tower at A is ϕ, Find ϕ and the height of the tower

Grade:11

1 Answers

Jitender Pal
askIITians Faculty 365 Points
6 years ago
Hello Student,
Please find the answer to your question
Let PQ be the tower of height h. A is in the north of O and P is towards east of A.
∴ ∠OAP = 90°
∠ QOP = 30°
∠QAP = ϕ
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∠ AOP = α s. t. tan α = 1/√2
Now in ∆OPQ, tan 30° = h/OP ⇒ OP = h/√3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
In ∆ APQ, tan ϕ = h/AP ⇒ AP = h cot ϕ . . . . . . . . . . . . . . . . . . . . (2)
Given that,
tan α = 1/√2 ⇒ sin α = tan α/√1 + tan2 α = 1/3
Now in ∆ AOP, sin α = AP/OP ⇒ 1/√3 = h cot ϕ/h √3 [ Using (1) and (2)]
⇒ cot ϕ = 1
⇒ ϕ = 45°
Again in ∆OAP, using Pythagoras theorem, we get
OP2 = OA2 + AP2
⇒ 3h2 = 90000 + h2 cot2 45°
⇒ h = 150 √2 m

Thanks
Jitender Pal
askIITians Faculty

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