The feet of two vertical poles of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a d...

Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.

Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with

The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:

(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]

(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]

(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where