Question

A 6 feet tall man sees an apple on the ground, $6\sqrt{3}$ feet away from him. What is the angle of depression when he is looking at the apple?

Answer 1

Imagine the man standing to be a vertical straight line, AB, and the position of the apple to be at a point C.
Now if we connect the man’s head and feet to the position of the apple we can imagine it to be a right triangle with vertices ABC, as in fig(i).
We already know that the height of the man is 6 feet and he is standing $6\sqrt{3}$ feet away from the apple. Therefore, we can say that, $AB=6$ feet and $BC=6\sqrt{3}$ feet.
The angle of depression is the angle the line of sight makes with the horizontal level, i.e. the angle AC and AD. Let us name the angle $\theta$, as in fig(ii).
$\angle DAB=\angle CBA={90}^{\circ }.$
Therefore we can say that $DA\parallel BC.$
Now, we can say that $\angle DAC=\angle ACB$, because they are alternate interior angles of the parallel lines $AD$ and $BC.$
Let’s apply trigonometric ratios in the right angled triangle $△ABC$ to find $\theta$.

$\tan \theta =\frac{AB}{BC}$
$=\frac{6}{6\sqrt{3}}$
$=\frac{1}{\sqrt{3}}$
We know that $\tan \left({30}^{\circ }\right)=\frac{1}{\sqrt{3}}$
$\theta ={30}^{\circ }$