1. ## Triangle problem

Hi all,
First post in the forum. Actually the first time asking for help online. Lets see how this works.

A small fire is sighted from ranger stations A and B. The bearing of the fire from A is N35E, and the bearing of the fire from B is N49W. Station A is 1.3 miles due west of station B.

I solved part a) (How far is the fire from each ranger station?) From station A it is 0.9 miles and from station B it is 1.1 miles.

b) At fire station C, which is 1.5 miles from A, there is a helicopter that can be used to drop water on the fire. If the bearing of C from A is S42E, find the distance from C to the fire, and find the bearing of the fire from C.

Tried really hard to figure out part b) without any luck! Help would be appreciated.

2. ## Re: Triangle problem

I hope that you recognise the angles. Otherwise, from this picture, can you get the answer for part (b)? The red parts are the ones you were asked to find.

You can get the length of line CF by using the cosine rule.

Take note that AF = 0.9915... mile = 0.99 miles and BF = 1.09 miles. Try keeping your answers to a reasonable precision.

3. ## Re: Triangle problem

Thanks! CF is 1.9 miles and the angle worked out to be 14.8 degrees. Let me know if there's a mistake.

4. ## Re: Triangle problem

I get 1.97439... which becomes 1.97 miles, or if you prefer 2 significant figures, it becomes 2.0 miles.

Okay, now could you show me how you got your angle? I got 12.75 somehow...

Well, I got angle AFC as 47.75 degrees and since angle AFS (S is south) is 35 degrees, the difference becomes 12.75 degrees.

I tried another method, finding angle ACF instead, giving me 29.25 degrees and since angle ACN (N is north) is 42 degrees the difference gives me 12.75 degrees again.

5. ## Re: Triangle problem

When you said your AFC is 47.75 degrees. I used the sine rule and found AFC to be 49.57 degrees.

The way I found the angle was:
ACN (N being north) is 42 degrees.
CAF is 103 degrees. [180 - (35+42) = 103]
Applying the sine rule to find ACF : $\displaystyle \frac{\sin103}{1.92}=\frac{\sin x}{0.9}$
therefore ACF = 27.18 degrees
ACN - ACF = 42 - 27.18 = 14.8 degrees