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magentarita · #3 November 20th, 2008, 07:09 AM

Quote:

Originally Posted by Soroban

Hello, magentarita!

Code:

                              L
                              *
                          * * |
                      *   *   |
                  *     *     | h
              *       *       |
          * 37°     * 50°     |
      * - - - - - * - - - - - *
      A    250    B     x     M

The lighthouse is: $LM = h$

The ship was at $A. \;\;\angle LAM = 37^o$

The ship moves to $B\!:\;AB = 250.\;\;\angle LBM = 50^o$

Let $x = BM.$

In $\Delta LMB\!:\;\;\tan50^o \:=\:\frac{h}{x} \quad\Rightarrow\quad h \:=\:x\tan50^o$ .[1]

In $\Delta LMA\!:\;\;\tan37^o \:=\:\frac{h}{x+250} \quad\Rightarrow\quad h \:=\:(x+250)\tan37^o$ .[2]

Equate [1] and [2]: . $x\tan50^o \:=\:(x+250)\tan37^o \quad\Rightarrow\quad x\tan50^o \:=\:x\tan37^o + 250\tan37^o$

. . $x\tan50^o - x\tan37^o \:=\:250\tan37^o \quad\Rightarrow\quad x(\tan50^o - \tan37^o) \:=\:250\tan37^o$

. . $x \;=\;\frac{250\tan37^o}{\tan50^o-\tan37^o} \;=\;429.914... \;\approx\;430$ feet

Therefore, the ship was about 330 feet from the rocks.

Your explanations makes me love math more each day.