From a window $\displaystyle 15$ metres high 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 $\displaystyle 30$° and $\displaystyle 45$° respectively. Find the height of the opposite house. (Use $\displaystyle \sqrt{3}=1 \cdot 732$ )

In right \(\displaystyle \Delta \mathrm{BPC}\), \(\displaystyle \tan 45^{\circ}=1=\frac{15}{x}\) \[\Rightarrow x=15 \] In right \(\displaystyle \Delta \mathrm{BPD}\), \(\displaystyle \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{x}\) \[\Rightarrow \quad \mathrm{h}=\frac{15 \sqrt{3}}{3} \text { or } 5 \sqrt{3} \] \[\begin{aligned} \therefore \text { Height of opposite house }(\mathrm{CD}) & =5 \sqrt{3}+15 \\ & =5(1 \cdot 732)+15=23 \cdot 66 \mathrm{~m} \end{aligned} \]