Question

An observer is 5.2 km from the launch pad. At a particular time, the angle of elevationfrom the observer to a vertically ascending missile is 31°. How high (nearest tenth km) isthe missile? Ignore the observer's height. Please draw a diagram

Answer 1

EXPLANATION

To solve the question, we will have to sketch the image

We are to use the sketch above to compute the height y

$tan31^0=\frac{opposite}{adjacent}=\frac{y}{5.2km}$

Then, we will make y the subject of the formula

$\begin{gathered} y=5.2\times\text{ tan31} \\ y=5.2km\times0.6009 \\ y=3.124km \end{gathered}$

Therefore, the value of the height will be 3.1km