Question

You sight a rock climber on a cliff at a 32 to the power of degree angle of elevation. Your eye level is 6 ft above the ground and you are 1000 ft from the base of the cliff. What is the approximate height of the rock climber from the ground. Round to the nearest foot.

Answer 1

The problem above is modelled in the diagram below

The distance of the climber from the ground is $x+6ft$, where 6 ft is the height of the observer from the ground

We use the trigonometry ratios to work out $x$

$tan(32)= \frac{x}{1000}$
$x=1000tan(32)$
$x=625 ft$ (to the nearest integer)

Hence, the height of the climber from the ground is 625+6 = 631 ft