Question

in the diagram below, a hiker at point $a$ is approaching a cliff. the vertical height of the cliff is 127 feet. the angle of elevation from point $a$ to the top of the cliff is $18^{\circ }$.
draw
after moving to point $b$, the angle of elevation becomes $23^{\circ }$. find, to the nearest foot, the distance traveled from point $a$ to point $b$.
show your work here

Explanation:

Step1: Find distance from A to cliff base

Let $x$ = distance from A to cliff base. Use $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$.
$\tan(18^\circ)=\frac{127}{x} \implies x=\frac{127}{\tan(18^\circ)}$
Calculate $\tan(18^\circ)\approx0.3249$, so $x\approx\frac{127}{0.3249}\approx390.9$ feet.

Step2: Find distance from B to cliff base

Let $y$ = distance from B to cliff base. Use $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$.
$\tan(23^\circ)=\frac{127}{y} \implies y=\frac{127}{\tan(23^\circ)}$
Calculate $\tan(23^\circ)\approx0.4245$, so $y\approx\frac{127}{0.4245}\approx299.2$ feet.

Step3: Compute distance AB

Subtract $y$ from $x$ to get $AB$.
$AB = x - y \approx 390.9 - 299.2$

Answer:

92 feet