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Soroban · #2 November 20th, 2008, 12:14 AM

Hello, magentarita!

Quote:

A ship captain sights the angle of elevation of 37° to the top of a lighthouse.
After traveling 250 feet toward the lighthouse, the new angle of elevation is 50°.
There are dangerous rocks 100 feet from the base of the lighthouse.
How close to the rocks is the ship at the time of the second sighting (nearest foor)?

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.

The following users thank Soroban for this useful post:
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