2. Problems involving angle of elevation

The altitude of the plane from point $C$ on the ground is $1500 \ m$. If the angle of elevation of the person from point $A$ is $60^{\circ}$, then find the distance of the person from point $A$ to $C$.

 

Solution:

 

 

Let $A$ be the position of the person, $B$ be the position of the airplane, and $C$ be the point.

 

Then, from the given data, we have:

 

$\theta = 60^{\circ}$ and $BC = 1500 \ m$

 

In the right triangle $ACB$, we have:

 

$tan \ \theta = \frac{BC}{AC}$

 

$tan \ 60^{\circ} = \frac{1500}{AC}$

 

$AC = \frac{1500}{tan \ 60^{\circ}}$

 

$AC = \frac{1500}{\sqrt{3}}$

 

$AC = 500\sqrt{3}$

 

Therefore, the distance between the points $A$ and $C$ is $500\sqrt{3} \ m$.