# Thread: Find the Length Of Track.

1. ## Find the Length Of Track.

Ok final one. I think, possibly I know how to do this.

Ok question goes sorta like this [summarized]

An incline that transports passengers back and forth [up and down] a mountain, has an angle elevation of 30 degrees. Angle of elevation to the top of the track from a point horizontal 100 ft from the base which is about 26.8 degrees. What is the lengh of the track?

Uhm, too lazy to draw the pretty triangle lol.

On this problem do i just do 180-30-26.8= 123.2

and carrry it on out by uaing the law of sines ( a / sinblah = b / sinblah )?

2. Hello, topaz192!

You have the right idea, but you need a better diagram.

An incline that transports passengers up and down a mountain,
has an angle elevation of 30°. From a point horizontal 100 ft from the base,
the angle of elevation to the top is 26.8°.
What is the lengh of the track?
Code:
                                      * B
* * |
*   *   |
*     *     |
*       *       |
*         *         |
*           *           |
* 26.8°  150° * 30°         |
* - - - - - - - * - - - - - - - *
D      100      A               C
We have: . $\angle BAC = 30^o \quad\Rightarrow\quad \angle BAD = 150^o$

. . and: . $DA = 100,\;\angle D = 26.8^o$

In $\Delta BAD,\;\angle DBA \;=\;180^o - 26.8^o - 150^o \;=\;3.2^o$

Law of Sines: . $\frac{AB}{\sin26.8^o} \:=\:\frac{100}{\sin3.2^o}$

Therefore: . $AB \;=\;\frac{100\sin26.8^o}{\sin3.2^o} \;=\;807.7129786$ feet.

3. Oh ok I sooo get it now! :]
At first I didnt know how you got 150. But I know now. [180-30]

Hmmm. No questions here. I finished working all the rest of the problems that relate to this.

Thank you for your time!!