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Thread: Trig Problems in 3D

  1. #1
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    Unhappy Trig Problems in 3D

    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

    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?
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    Quote Originally Posted by SOHCAHTOA View Post
    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

    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 $\displaystyle a$ = horizontal distance from mast to buoy #1

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

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

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

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

    let $\displaystyle c$ = distance between the two buoys

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