# Trig Word Problem, Need Help

• Nov 18th 2008, 01:54 PM
Vivin Spinach
Trig Word Problem, Need Help
Two spotlights, one blue and the other white, are placed 6.0m apart on a track on the ceiling of a ballroom. A stationary observer standing on the ballroom floor notices that the angle of elevation is 45 degrees to the blue spotlight and 70 degrees to the white one. How high, to the nearest tenth of a metre, is the ceiling of the ballroom?

I know that the answer is 4.4m, because I looked it up in the back of the book, but I have no idea how they got that. :S

If someone could help me out that would be great. I am in disparate need of it.
• Nov 18th 2008, 08:43 PM
Soroban
Hello, Vivin Spinach!

I don't agree with their answer . . .

Quote:

Two spotlights, one blue and the other white, are placed 6.0m apart on a track
on the ceiling of a ballroom.
An observer on the floor sees that the angle of elevation is 45° to the blue spotlight
and 70° to the white one.
How high, to the nearest tenth of a metre, is the ceiling of the ballroom?

Code:

                  W    6    B       C - - - - - * - - - - * -             70° * 110°  * |                 *      *  |               *      *    |               *    *      |             *    *        | h             *25°*          |           *  *            |           * *              |         ** 45°            |       O * - - - - - - - - - * A

The white light is at $\displaystyle W$, the blue light at $\displaystyle B.\;\;WB = 6$

The observer is at $\displaystyle O.\;\;\angle BOA = 45^o,\;\;\angle WOA = 70^o \quad\Rightarrow\quad \angle WOB = 25^o$
. . We also have: .$\displaystyle \angle CWO = 70^o \quad\Rightarrow\quad \angle BWO = 110^o$
Let $\displaystyle h = AB$

In $\displaystyle \Delta BWO$, use the Law of Sines:

. . $\displaystyle \frac{OB}{\sin110^o} \:=\:\frac{6}{\sin25^o} \quad\Rightarrow\quad OB \:=\:\frac{6\sin110^o}{\sin25^o} \:\approx\:13.34$

In $\displaystyle \Delta BAO\!:\;\;\sin45^o \:=\:\frac{h}{OB} \quad\Rightarrow\quad h \:=\:OB\sin45^o \:=\:13.34\sin45^o$

Therefore: .$\displaystyle OB \:\approx 9.4\text{ m}$

• Nov 19th 2008, 09:59 AM
masters
Quote:

Originally Posted by Soroban
Hello, Vivin Spinach!

I don't agree with their answer . . .

Therefore: .$\displaystyle {\color{red}h} \:\approx 9.4\text{ m}$

Soroban, I worked this one as well and found the same answer you did. I abandoned my solution thinking I had missed something obvious. Oh, I corrected your typo at the end there.