(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the t...
Question 1 Report
(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.
(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].
Answer Details
Let the horizontal distance between the building and the foot \(F\) of the tower be \(x\), and the tower height be \(H\). \(P\) is at the foot of the building, \(Q\) is its top, \(20\text{ m}\) directly above \(P\).
From \(P\): \(\tan 65° = \dfrac{H}{x}\), so \(H = x\tan 65°\).
From \(Q\) (height 20 m, same horizontal distance \(x\)): \(\tan 25° = \dfrac{H - 20}{x}\), so \(H - 20 = x\tan 25°\).
Subtracting: \(20 = x(\tan 65° - \tan 25°) = x(2.1445 - 0.4663) = 1.6782x\).
\(x = \dfrac{20}{1.6782} = 11.92\text{ m}\).
(a) Distance PT: \(PT\) is the line of sight from \(P\) to \(T\), at elevation \(65°\):
\(PT = \dfrac{x}{\cos 65°} = \dfrac{11.92}{0.4226} = 28.20 \approx \mathbf{28.2\text{ m}}\) (3 s.f.).
(b) Height of tower: \(H = x\tan 65° = 11.92 \times 2.1445 = 25.56 \approx \mathbf{25.6\text{ m}}\) (3 s.f.).
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