A window of a house is \(h \ \text{metres}\) above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be \(\alpha\) and \(\beta\), respectively. Prove that the height of the other house is \(h(1 + tan \ \alpha \ cot \ \beta)\) \(\text{metres}\).
 
 Answer:
 
 In \(\triangle QWM\), we have:
 
\(tan \ \alpha =\)
 
\(tan \ \alpha =\)
 
\(x=\)
 
In \(\triangle OWM\), we have:
 
\(tan \ \beta=\)
 
\(tan \ \beta=\)
 
\(x=\)  
 
By equating and Simplifying the values of \(x\) we get
 
\(H\) = h(1+tanαcotβ)
 
Hence proved
Answer variants:
Hhtanα
htanβ
QMWM
hx
OMWM
Hhx