A tower stands vertically on the ground. From a point on the ground, 20 m away from the foot of the tower, the angle of elevation of the top the tower is 60°.
What is the height of the tower?
Solution:
A common notion in trigonometry, specifically, is the angle of elevation, which has to do with height and distance. It is described as an angle formed by the horizontal plane and an oblique line between the observer's eye and a target above it.
Given:
Distance between the foot of the tower and point of observation = 20 m = BC
Angle of elevation of the top of the tower = 60° = θ
And, Height of tower (H) = AB
Now, from fig. ABC
ΔABC is a right angle triangle,
So,
\begin{aligned} &\tan \theta=\frac{\text { Opposite side }(\mathrm{AB})}{\text { Adjacent side }(\mathrm{BC})} \\ &\begin{array}{l} \Rightarrow \tan \mathrm{C}=\frac{\mathrm{AB}}{\mathrm{BC}} \\ \mathrm{AB}=20 \tan 60^{\circ} \\ \mathrm{H}=20 \times \sqrt{3} \\ \quad=20 \sqrt{3} \end{array} \\ &\therefore \text { Height of the tower }=20 \sqrt{3} \mathrm{~m} \end{aligned}
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