From a window (\(h\) metres high above the ground) of a house 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 \(\theta_1\) and \(\theta_2\) respectively. Demonstrate that the height of the opposite house is \(h\left(1 + \frac{cot \ \theta_2}{cot \ \theta_1}\right)\).
Proof:
Let \(W\) be the point of the window. Let \(PQ\) be the house on the opposite side.
Then, \(WA\) is the width of the street.
Height of the window \(= h \ m\) \(= WR = AQ\)
Let us assume that \(PA = x\)
In the right \(\triangle PAW\), \(tan \ \theta_1 = \frac{PA}{AW}\)
---- (\(1\))
In the right \(\triangle QAW\), \(tan \ \theta_2 = \frac{AQ}{AW}\)
---- (\(2\))
From equations (\(1\)) and (\(2\)), we get:
Height of the opposite house \(= PA + AQ\)
Therefore, the height of the opposite house is \(h\left(1 + \frac{cot \ \theta_2}{cot \ \theta_1}\right)\).
Answer variants:
\(= \frac{h \ cot \ \theta_2}{cot \ \theta_1} + h\)
\(tan \ \theta_2 = \frac{h}{AW}\)
\(x = \frac{h \ cot \ \theta_2}{cot \ \theta_1}\)
\(x \ cot \ \theta_1 = h \ cot \ \theta_2\)
\(= h\left(1 + \frac{cot \ \theta_2}{cot \ \theta_1}\right)\)
\(AW = h \ cot \ \theta_2\)
\(tan \ \theta_1 = \frac{x}{AW}\)
\(AW = \frac{h}{tan \ \theta_2}\)
\(AW = \frac{x}{tan \ \theta_1}\)
\(AW = x \ cot \ \theta_1\)