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Grade 11Algebra

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

Radhika Batra

12 Years agoGrade 11

1 Answer

Jitender Pal

12 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

∠ 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
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