Trig Problems in 3D

Quote:

Originally Posted by SOHCAHTOA

I'm currently learning this in math...but I have no clue on how to start the question! This is a 3D trigonometry question so I know it's best to always draw a diagram first. Problem is, I don't how where to begin (Crying)

The Crows-nest of a yacht is 50.0 m above water level. The angle of depression from the crows nest to a buoy due west of the boat is 40/degree/. The angle of depression from another buoy S70/degress/W of the yacht is 34/degrees/. How far apart are the buoys?

The answer is 27.3m...but how did they do it?

let $a$ = horizontal distance from mast to buoy #1

$b$ = horizontal distance from mast to buoy #2

$a = \frac{50}{\tan(40)}$

$b = \frac{50}{\tan(34)}$

angle on the water between $a$ and $b$ is 20 degrees

let $c$ = distance between the two buoys

$c = \sqrt{a^2 + b^2 - 2ab\cos(20)}$