From a window x meters high above the ground in a street, the angles of elevation and depression of the top and the foot of the other house on the opposite side of the street  are  α  and  β  respectively. Show  that  the  height  of  the  opposite house is x(1+ tanα cotβ)Meters.

Ans:    Let AB be the house and P be the window

Let BQ = x ∴ PC = x

Let AC = a

In Δ PQB, tan θ = PQ/QB or tan θ = h/x

∴ x = h/tan θ    = h cot θ

In Δ PAC, tan θ = AC/PC or tan θ = a/x

∴ a = x tan θ > (h cot θ) tan θ = h tan θ cot θ.

∴ the height of the tower = AB = AC + BC

= a + h = h tan θ cot θ + h = h (tanθ cot θ + 1)