Question

An observer (O) is located 500 feet from the school (S). The observer notices a bird (B) flying at a 39 degree angle of elevation from his line of sight. How high is the bird flying over the school? Pls show work

Answer 1

To answer this question we will use the following diagram as a reference:

Recall that in a right triangle we get:

$\tan\theta=\frac{OppositeLeg}{AdjacentLeg}.$

From the given diagram we get:

$\tan39º=\frac{SB}{OS}=\frac{SB}{500ft.}$

Therefore:

$SB=500ft\tan39º.$

Simplifying the above result we get:

$SB\approx404.89ft.$

Answer: 404.89ft.