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Two poles X and Y stand 3√2 m apart, such that the angle of elevation of the top of pole X as observed from the foot of pole Y is 30° and the angle of elevation of the top of pole Y as observed from the top of pole X is 60°. What is the angle of elevation of the top of pole Y from the foot of pole X?

Two poles X and Y stand 3√2 m apart, such that the angle of elevation of the top of pole X as observed from the foot of pole Y is 30° and the angle of elevation of the top of pole Y as observed from the top of pole X is 60°. What is the angle of elevation of the top of pole Y from the foot of pole X? Correct Answer tan<sup>- 1</sup> (4/√3)

Given:

distance between poles = XY = AC = 3√2 m

Formula Used:

tan θ = perpendicular/base

Calculation:

Let A and B be the top of poles X and Y respectively, as shown below,

Let θ be the angle of elevation of top of pole Y as observed from foot of pole X,

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Given, distance between poles = XY = AC = 3√2 m

Considering ∆AXY,

⇒ tan 30° = perpendicular/base = AX/XY

⇒ 1/√3 = AX/3√2

⇒ AX = 3√2/√3 = √6 m

Considering ∆ ACB,

⇒ tan 60° = perpendicular/base = BC/AC

⇒ √3 = BC/3√2

⇒ BC = 3√6 m

⇒ BY = BC + CY = BC + AX = 3√6 + √6 = 4√6 m

Now,

Considering ∆XYB,

⇒ tan θ = perpendicular/base = BY/XY

⇒ tan θ = 4√6/3√2

⇒ tan θ = 4/√3

∴ Required angle = θ = tan- 1 (4/√3)

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