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 α and respectively. Prove that the height of the other house is h(1+tan α cot) metres.

Let WB = h m, height of the smaller window & QO = H m, height of the other window.

Let BO = x m, distance between the two windows.

∠QWM = α & ∠WOB = β

In ∆WOB,

In ∆QWM,

Comparing eq. 1 and eq. 2, we get

(H – h)/tan α = h/tan β

⇒ (H – h) tan β = h tan α

⇒ H tan β – h tan β = h tan α

⇒ H tan β = h tan α +h tan β

⇒ H tan β = h (tan α + tan β)

⇒ H = (h (tan α + tan β))/tan β

⇒ H = h (tan α/tan β + 1)

⇒ H = h (1+tan α/tan β)

⇒ H = h (1+tan α cot β)

Hence, proved.