Question

A helicopter is sighted at the same time by two ground observers who are 5 miles apart, both ahead of the helicopter and in its direction of travel. They report angles of elevation of 19.5° and 31.2° to the helicopter. Find the distance from the helicopter to the further observer, to the nearest tenth of a mile.

Answer 1

Answer:

The distance from the helicopter to the further observer is:

d = 12.77 miles

Step-by-step explanation:

The observer with 19.5° of elevation will be denoted by 1

The observer with 31.2° of elevation will be denoted by 2.

Here we have two right triangles. Both of them have the same opposite sides, so will be useful to use the tangent function.

For the second observer:

$tan(31.2)=\frac{h}{x}$ (1)

Where:

• h is the opposite side
• x is the adjacent side (distance from the observer 2 to the point where the angle of elevation is 90° )

For the first observer:

$tan(19.5)=\frac{h}{5+x}$ (2)

Where:

• h is the opposite side
• 5+x is the adjacent side (distance from the observer 1 to the 90° elevation point, 5 miles is the distance between observes )

Let's find x from equations (1) and (2). Both of them have the variable "h" so we can equal these equations.

$xtan(31.2)=(5+x)tan(19.5)$

$x(tan(31.2)-tan(19.5))=5tan(19.5)$

$x=\frac{5tan(19.5)}{tan(31.2)-tan(19.5)}$

$x=7.04\: miles$

Now, if we use the right triangle with 19.5° of elevation, the hypotenuse will be the distance from the helicopter and the further observer (let's call d).

Let's use the cosine function to find d.

$cos(19.5^{\cric})=\frac{5+x}{d}$

$d=\frac{5+7.04}{cos(19.5^{\cric})}$

$d=12.77\: miles$

I hope it helps you!