# Word Problem

• Jan 28th 2013, 03:32 PM
sakonpure6
Word Problem
The crows-nest of a yacht is 50.0m above the water level. The angle of depression from the crows nest to a buoy due West of the boat is 40 degrees. The angle of depression to another buoy is South 70 degrees West of the yacht is 34 degrees. How far apart are the buoys? (Answer 27.3m)

A diagram is really appreciated!!
• Jan 28th 2013, 05:29 PM
Soroban
Re: Word Problem
Hello, sakonpure6!

Quote:

$\text{The crows-nest of a yacht is 50.0 m above the water level.}$
$\text{The angle of depression from the crows-nest to a buoy }A\text{ due west of the boat is }40^o.$
$\text{The angle of depression to another buoy }B\text{ at }S\,70^o\,W\text{ of the yacht is }34^o.$
$\text{How far apart are the buoys? \;(Answer 27.3m)}$

Looking toward buoy A, we have this diagram:
Code:

          - - - - * C             40o * *               *50o*             *    * 50           *      *         *        *     A *  *  *  *  * N           x
We have: . $\tan50^o \:=\:\frac{x}{50} \quad\Rightarrow\quad x \:=\:50\tan50^o$
. . $x \:\approx\:59.588$

Looking toward buoy B, we have this diagram.
Code:

          - - - - * C             34o * *               *56o*             *    * 50           *      *         *        *     B *  *  *  *  * N             y
We have: . $\tan56^o \:=\:\frac{y}{50} \quad\Rightarrow\quad y \:=\:50\tan56^o$
. . $y \:\approx\:74.13$

Looking down, we have this diagram:
Code:

            59.59     A o  *  *  *  *  *  o C       *          20o * :         *          *70o:         *        *    :           *    * 74.13 :           *  *        :             o             B
Law of Cosines:

. . $AB^2 \;=\;59.59^2 + 74.13^2 - 2(59.59)(74.13)\cos20^o \;=\;744.216042$

. . $AB \;=\;27.28032335 \;\approx\; 27.3\text{ m}$
• Jan 28th 2013, 07:03 PM
sakonpure6
Re: Word Problem
O great thank you!