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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

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EXPLANATION

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

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

[tex]tan31^0=\frac{opposite}{adjacent}=\frac{y}{5.2km}[/tex]

Then, we will make y the subject of the formula

[tex]\begin{gathered} y=5.2\times\text{ tan31} \\ y=5.2km\times0.6009 \\ y=3.124km \end{gathered}[/tex]

Therefore, the value of the height will be 3.1km

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