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 distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30°. Calculate the :
(i) distance RT ; (ii) distance of the foot of the taller pole from P, correct to three significant figures ; (iii) angle of elevation of T from P, correct to one decimal place.
Explanation
(i) From the figure,
(sin 30° = frac{4}{RT} implies RT = frac{4}{sin 30°})
= (frac{4}{0.5} = 8m)
(ii) |RQ| = |XY|
(therefore |XY|^{2} = 8^{2} – 4^{2} = 64 – 16 = 48)
(|XY| = sqrt{48} = 4sqrt{3} m = 6.928m)
But |PY| = |PX| + |XY|
= 20 + 6.928 = 26.928m (approxeq) 26.9m.
(iii) Let the angle of elevation = (theta)
(tan theta = frac{TY}{PY} = frac{7}{26.928})
(tan theta = 0.26 implies theta = tan^{-1} (0.26))
(theta = 14.57° approxeq 14.6°) (to 1 decimal place)