# Thread: I know this is simple but...

1. ## I know this is simple but...

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

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

2. ## Re: I know this is simple but...

Hello, Jaydenaus!

A person on a boat 2 meters above sea level, has an angle of elevation of 16o to the top of a lighthouse.
After moving 50 meters directly away from the lighthouse, the angle of elevation is 12o.
Find the height of the top of the lighthouse above sea level.
Code:
                              *A
** |
* *   |
*  *     |
*   *       |h
*    *         |
*     *           |
* 12o  * 16o          |
D* - - - * - - - - - - - *B
2|  50   C       x       |2
F* - - - - - - - - - - - *E
The height of the lighthouse above sea level is $\displaystyle AE = h+2.$
$\displaystyle \angle ACB = 16^o,\;\angle ADB = 12^o,\;BC = x,\;CD = 50$

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

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

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

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

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

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

3. ## Re: I know this is simple but...

Thanks a lot, answered my question completely. =)