Question

A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be alpha and beta, respectively.

Prove that the height of the other house is = h(1 + tanα tanβ)

Answer 1

I think answer should be

I think answer should be height of another house is h(1 + tanα.cotβ)

see attached figure,

Let AE is the window of one house which is h m metre above from the ground . from the window , angle of elevation (α ) and angle of depression ( β ) of the top and bottom of another house of height BD is shown in figure.

now, AE = BC = h ---------(1)

from ∆BCE ,

tanβ = BC/CE = h/CE

CE = h/tanβ -------(2)

from ∆DEC ,

tanα = DC/CE

from equation (1) CE = h/tanβ = hcotβ, put it here,

tanα = DC/hcotβ

DC = h.tanα.cotβ --------(3)

now, we can see that height of another house is BD

BD = BC + CD

from equations (1) and (3)

BD = h + h.tanα.cotβ = h(1 + tanα.cotβ)

$hence, \bold{BD=h(1+tan{\alpha}.cot{\beta})}$

Answer 2

In ΔACD,
CD/AD=tanβ
h/AD=tanβ⇒h=AD×tanβ ⇒AD=h/tanβ
In ΔAED
DE/AD=tan α
DE/(h/tan β)=tan α
DE=tanα×(h/tanβ)=(tan α×h)/tan β
CE=CD+DE
=h+[(tan α×h)/tanβ]=h[1+(tan α/tan β)]
Height of the other house=h[1+(tanα/tanβ)]