(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, cal...

Answer Details

(a) Let the building have height \(h\) and foot \(B\), with \(P\) on the ground \(50\ \text{m}\) from \(B\). The angle of depression from the top \(T\) to \(P\) equals the angle of elevation from \(P\) to \(T\) (alternate angles), so this angle at \(P\) is \(23.6^\circ\).

\[\tan 23.6^\circ = \frac{h}{50}\]\[h = 50 \times \tan 23.6^\circ = 50 \times 0.4369 = 21.85\ \text{m}\]

Height \(\approx 22\ \text{m}\) (to the nearest metre).

(b) From the diagram, \(PT \parallel SU\) and \(QS \parallel TR\), with \(S\), \(R\), \(U\) on the bottom line where \(|SR| = 6\ \text{cm}\) and \(|RU| = 10\ \text{cm}\).

Find the height between the parallel lines. Triangle \(TRU\) has base \(RU = 10\ \text{cm}\) on the bottom line, and apex \(T\) on the top line, so its height is the perpendicular distance \(H\) between \(PT\) and \(SU\):

\[\text{Area of } \Delta TRU = \frac{1}{2} \times RU \times H = 45\]\[\frac{1}{2} \times 10 \times H = 45 \Rightarrow H = 9\ \text{cm}\]

Find \(QT\). Since \(QT \parallel SR\) (both on the parallel top and bottom lines) and \(QS \parallel TR\), the figure \(QTRS\) is a parallelogram, so:

\[QT = SR = 6\ \text{cm}\]

Length of \(SU\):

\[SU = SR + RU = 6 + 10 = 16\ \text{cm}\]

Area of trapezium \(QTUS\) (parallel sides \(QT = 6\ \text{cm}\) and \(SU = 16\ \text{cm}\), height \(H = 9\ \text{cm}\)):

\[\text{Area} = \frac{1}{2}(QT + SU) \times H = \frac{1}{2}(6 + 16) \times 9\]\[= \frac{1}{2} \times 22 \times 9 = 99\ \text{cm}^2\]

Area of trapezium \(QTUS = 99\ \text{cm}^2\)