# A helicopter is flying at an altitude of 1200°m. The angle of elevation from the tower on the ground to the plane measures 36°. How far is the building to the plane?

Mar 31, 2018

$\text{Distance between the plane and the building is " = color(blue)(2041.56 " m}$

#### Explanation:

Given $B C = \text{ Altitude " = 1200 m, " Angle of elevation } = {36}^{\circ}$

To find $\text{Distance AC }$

ABC is a right triangle with AC the hypotenuse. Applying Pythagoras theorem,

$A C = x = \frac{B C}{\sin} X = \frac{1200}{\sin} 36 = 2041.56 \text{ m}$

Mar 31, 2018

#### Explanation:

Assumptions:

1. Angle of elevation from the tower on the ground to the plane: $\Rightarrow$ angle of elevation is from the foot of the tower $\angle O F H = {36}^{\circ}$.

2. How far is the building to the plane?: $\Rightarrow$ horizontal distance from the foot of the tower $\overline{O F}$.

Please refer to the diagram (NOT drawn to scale) demonstrating the assumptions used:

$\overline{O H} = 1200$ meters

$\angle H O F = {90}^{\circ}$

$\angle O F H = {36}^{\circ}$

We must find the magnitude of $\overline{O F}$.

For the $\angle O F H$,

$\overline{O H}$ is the Opposite Side.

$\overline{O F}$ is the Adjacent Side.

Since we must find the magnitude of $\overline{O F}$, we use the reatio for $\tan \left(\theta\right)$

$\tan \left(\theta\right) = \tan \left({36}^{\circ}\right) =$ Opposite Side$/$Adjacent Side

$\Rightarrow \tan \left({36}^{\circ}\right) = \frac{O H}{O T}$

$\Rightarrow \tan \left({36}^{\circ}\right) = \frac{1200}{O T}$

$\Rightarrow O T \cdot \tan \left({36}^{\circ}\right) = 1200$

$\Rightarrow O T = \frac{1200}{\tan} \left({36}^{\circ}\right)$

Using the calculator, $\tan \left({36}^{\circ}\right) \approx 0.72654$

$\Rightarrow O T = \frac{1200}{0.72654} \approx 1651.664$

Hence, $\overline{O T} \approx 1651.67$ meters.

I have used a software to construct the problem for visual verification:

Scale Factor : 1 unit on the graph = 100 meters

All calculations are carried out using available functions in the software.