# Thread: Ship Near A Rock

1. ## Ship Near A Rock

A ship captain at sea uses a sextant to sight an angle of elevation of 37°to the top of a lighthouse. After the ship travels 250 feet directly towardthe lighthouse, another sighting is made, and the new angle of elevation
is 50°. The ship’s charts show that there are dangerous rocks 100 feet from the base of the lighthouse. Find, to the
nearest foot, how close to the rocks the ship is at the time of the second sighting.

2. Hello, magentarita!

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.

3. ## ok.........

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" alt="\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" alt="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.