A window of a house in h meter above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be $α$ and $β$ respectively. Prove that the height of the house is h(1+tan~\alpha~tan~\beta) metres.

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Solution

Let us observe the following figure.

The point of observation is at A.

The height of the window is h meters.

Thus, AB = CD = h meters

The top and bottom of another house in the opposite lane is E and C.

Therefore the angles of elevation and the depression are

$∠ E A D = α a n d ∠ D A C = β$

$C o n s i d e r △ A D E : t a n α = \frac{E D}{D A} => E D =$

$A D t a n α - - (1)$

$N o w c o n s i d e r △ A B C : t a n β = \frac{A B}{B C}$

$=> t a n β = \frac{h}{B C} => B C = h t a n β - - (2)$

From the figure, it is clear that AD = BC.

Thus, equation (2) becomes,

$A D = h c o t β$ -----(3)

Substitute the value of AD from equation (3) in equation (1), we have

$E D = A D t a n α = (h c o t β) t a n α$ -----(4)

The height of the house is CE

That is,

CE = CD + DE = h + DE ---(5)

Substitute the value of ED from equation (4) in equation (5), we hve

$C E = h + D E = h + h t a n α c o t β = h (1 + t a n α c o t β)$

Thus, the toltal height of the the housse is,

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A window of a house in h meter above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be α and β respectively. Prove that the height of the house is h(1+tan~\alpha~tan~\beta) metres.