From a window 15 meter high …

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 angles of deviation of the top D of house CD as seen from P is of 30^o and the angle of depression of the foot C of house CD as seen from P is of 45^o.

Let h metres be the height of the house CD.
We have
QD = CD- CQ = CD -A P = ( h - 15) metres.
In {tex}\triangle P Q C,{/tex} we have
tan 45^o = {tex}\frac { Q C } { P Q } \Rightarrow 1 = \frac { 15 } { P Q } \Rightarrow P Q = 15{/tex} metres.
In {tex}\triangle P Q C,{/tex} we have
{tex}\tan 30 ^ { \circ } = \frac { Q D } { P Q }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { \sqrt { 3 } } = \frac { h - 15 } { 15 } \Rightarrow h - 15 = \frac { 15 } { \sqrt { 3 } } \Rightarrow h - 15 = 5 \sqrt { 3 }{/tex}
{tex}\Rightarrow \quad h = 15 + 5 \times 1.732 = 23.66{/tex}metres.
Hence, the height of the opposite house is 23.66 metres.