Question

From a window 15 meters heigh above the ground in a street , the angles of elevation and depression of the top and the foot of another house on the opposite side of the street are ${30}^{\circ }and{45}^{\circ }$ respectively show that the height of the opposite house is 23.66 meters. (Take $\sqrt{3}$ = 1.732)

Answer 1

Let the window be at P at a height of 15 metres above the ground and CD be the house on the opposite side of the street such that the angle of devation of the top D of house CD as seen from P is of ${30}^{\circ }$ and the angle of depression of the foot C of house CD as seen from P is of ${45}^{\circ }$.
Let h metres be the height of the house CD.
We have,
QD = CD - CQ = CD - AP =$(h-15)$ metres.
In $\Delta$ PQC, we have
$\tan {45}^{\circ }$ = $\frac{QC}{PQ}\Rightarrow 1=\frac{15}{PQ}\Rightarrow PQ=15$ metres.
In $\Delta$ PQD, we have
$tan{30}^{\circ }=\frac{QD}{PQ}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-15}{15}\Rightarrow h-15=\frac{15}{\sqrt{3}}\Rightarrow h-15=5\sqrt{3}$
$Rightarrow$ h = 15 + 5 $\times$ 1.732 = 23.66 metres,
Hence, the height of the opposite house is 23.66 metres.