Question

Question 14
An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer 1

Let the angle of elevation of the top of the tower from the eye of the observer is $\theta$.

Given that, AB = 22 m, PQ = 1.5 m = MB

And, QB = PM = 20.5 m

$\Rightarrow AM=AB-MB$

= 22 - 1.5 = 20.5 m

Now, in $\Delta APM$, $tan\theta =\frac{AM}{PM}=\frac{20.5}{20.5}=1$

$\Rightarrow tan\theta =tan{45}^{\circ }$

Hence, required angle of elevation of the top of the tower from the eye of the observer is ${45}^{\circ }$.