I know this is simple but...

**Jaydenaus** · March 9th 2013, 05:03 PM

Just started math class again this year, and all prior knowledge went down the drain on holidays.=P

I know this is simple but...-img_20130310_100035.jpg

P.S. Glad to have joined MHF Community. =)

**Soroban** · March 9th 2013, 06:49 PM

Hello, Jaydenaus!

A person on a boat 2 meters above sea level, has an angle of elevation of 16^o to the top of a lighthouse.
After moving 50 meters directly away from the lighthouse, the angle of elevation is 12^o.
Find the height of the top of the lighthouse above sea level.

Code:

                              *A
                           ** |
                        * *   |
                     *  *     |
                  *   *       |h
               *    *         |
            *     *           |
         * 12^o  * 16^o          |
     D* - - - * - - - - - - - *B
     2|  50   C       x       |2
     F* - - - - - - - - - - - *E

The height of the lighthouse above sea level is $AE = h+2.$
$\angle ACB = 16^o,\;\angle ADB = 12^o,\;BC = x,\;CD = 50$

In $\Delta ABC\!:\;\tan16^o = \frac{h}{x} \quad\Rightarrow\quad x = \frac{h}{\tan16^o}\;\;[1]$
In $\Delta ADB\!:\;\tan12^o = \frac{h}{x+50} \quad\Rightarrow\quad x = \frac{h}{\tan16^o} - 50\;\;[2]$

Equate [2] and [1]: . $\frac{h}{\tan12^o} - 50 \:=\:\frac{h}{\tan6^o}$

n . . . . . . . . . . $\frac{h}{\tan12^o} - \frac{h}{\tan16^o} \:=\:50$

. . . . . . . . $h\left(\frac{\tan16^o-\tan12^o}{\tan12^o\tan16^o}\right) \:=\:50$

Hence: . $h \;=\;\frac{50\tan12^o\tan16^o}{\tan16^o-\tan112^o} \;=\;41.07735499 \;\approx\;41\text{ m}$

Answer: . $41 + 2 \;=\;43\text{ m}$

**Jaydenaus** · March 9th 2013, 08:21 PM

Thanks a lot, answered my question completely. =)

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