Question

A shooter standing on the ground horizontally 15 meter away from atree sees a bird on a branch at an altitude of 22 meters. If the eyes of theshooter are 1.5 meter above the ground level, what is the angle ofelevation?Round your answer to the nearest tenth of a degree.

Answer 1

Given: A man is standing at a distance of 15 m from a tree whose height is 22m.Now he sees a bird on th tree. The eye is 1.5 m above ground level, we have the find the angle of this elevation.

Let us draw the figure,

Now, we have to find the angle of elevation that is angle BCA.

Using trigonometric ratios ,we can easily find this angle,

$\begin{gathered} \text{tan(}\angle\text{BCE)}=\frac{perpendicular}{\text{base}} \\ =\frac{20.5}{15} \end{gathered}$

Here, we have taken the perpendicular as 22-1.5=20.5m

$\tan (\angle BCE)=1.3666$

So,

$\begin{gathered} \angle BAC=\tan ^{-1}(1.3666) \\ \angle BAC=53.8^{\circ} \end{gathered}$

This is the required angle of elevation,that is, 53.8 degrees.