Discussion
The elevation of T, from A is xº where tan xº = 2/5 and AF = 200 m. The elevation of T from B, where AB = 80 m is yº. Calculate : (a)the height of the tower TF, (b)the angle ’y’ correct to nearest degree
Answer: B
F is the foot of the tower, A and B are on the same straight line from it, with \( AF = 200 \) m and \( AB = 80 \) m, so \( BF = 120 \) m.
(a) Height of the tower. From A:
\[ \tan x^{\circ} = \frac{TF}{AF} = \frac{2}{5} \quad\Longrightarrow\quad TF = 200 \times \frac{2}{5} = 80 \text{ m} \]
(b) The angle y. From B, the same 80 m tower is only 120 m away:
\[ \tan y^{\circ} = \frac{80}{120} = \frac{2}{3} = 0.6667 \]
\[ y = \tan^{-1}(0.6667) = 33.69^{\circ} \approx 34^{\circ} \]
Moving 80 m closer raises the angle from about \( 21.8^{\circ} \) to \( 34^{\circ} \) - the elevation grows faster the nearer you get.
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