# Thread: i need help fast....x

1. ## i need help fast....x

im really stuck on this question could some one plz help me solve it.

An observer sees a ship from the top of a cliff of height 100 m. The ship is seen
through an instrument which measured the angle of depression to be 80
°. The observer
then moved back a distance of 10 m, at which point the ship and the top of the cliff

were in line. Find the distance from the base of the cliff to the ship.

thanx
xxx

2. Hello, Steph!

A challenging problem . . .
. . and I got some strange answers.

An observer sees a ship from the top of a cliff of height 100 m.
The ship is seen through an instrument which measured the angle of depression to be 80°.
The observer then moved back a distance of 10 m, at which point the ship
and the top of the cliff were in line.
Find the distance from the base of the cliff to the ship.

I assume that the instrument has a height $\displaystyle h.$
Code:
           A* - - - - - - - - - - - - E
| *  80°
h |   *
|     *
B*       *
|         *
|           *
|             *
100 |               *
|                 *
|                   *
|                     *
C* - - - - - - - - - - - * D
x

The instrument is: $\displaystyle AB = h$
The cliff is: $\displaystyle BC = 100$
The ship is at $\displaystyle D\!:\;CD = x$
$\displaystyle \angle EAD = 80^o$
In right triangle $\displaystyle AED\!:\;\;\tan80^o \:=\:\frac{h+100}{x}\quad\Rightarrow\quad h \:=\:x\tan80^o -100\;\;{\color{blue}[1]}$
Code:
    A'       A
*       *
| *     |
h |   *   |
|     * |
B'* - - - *B
10   |  *
|     *
|        *
100 |           *
|              *
|                 *
|                    *
C* - - - - - - - - - - - *D
x

The observer (and the instrument) moves 10 m to $\displaystyle A'B'.$
. . and $\displaystyle A',\,B,\,D$ are collinear.

Since $\displaystyle \Delta A'B'B \sim \Delta BCD\!:\;\;\frac{h}{10} \:=\:\frac{100}{x}\quad\Rightarrow\quad h \:=\:\frac{1000}{x}\;\;{\color{blue}[2]}$

Equate [1] and [2]: .$\displaystyle x\tan80^o - 100 \:=\:\frac{1000}{x} \quad\Rightarrow\quad x^2\tan80^o - 100x - 1000 \:=\:0$

Quadratic Formula: .$\displaystyle x \;=\;\frac{100 \pm\sqrt{100^2 + 4(\tan80^o)(1000)}}{2\tan80^o}$

The positive root is: .$\displaystyle x \;\approx\;\boxed{24.755\text{ meters}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Substitute into [1]: .$\displaystyle h \;=\;24.755\tan80^o - 100 \;\approx40.4$ m ??

Let me understand this . . .

The ship is less than 25 meters from the cliff,
. . but the instrument is over 130 feet high ??