# Thread: Trig class - bearing problem

1. ## Trig class - bearing problem

Please, help me to solve the problem:
Two fire towers are located 85 km apart on hills A and B. The bearing from A to B is north east. A fire F is observed from tower A at N10E and from B at N75W. The town at point C is on bearing of N25E from A and S70W from B. The observers report the wind is blowing the fire directly toward C at a rate 9km/h. How many hours do the officials have to evacuate the town?

I can't find the distance FC
Thank you.

2. Originally Posted by Romanka
Please, help me to solve the problem:
Two fire towers are located 85 km apart on hills A and B. The bearing from A to B is north east. A fire F is observed from tower A at N10E and from B at N75W. The town at point C is on bearing of N25E from A and S70W from B. The observers report the wind is blowing the fire directly toward C at a rate 9km/h. How many hours do the officials have to evacuate the town?

I can't find the distance FC
Thank you.
I've attached a sketch.

1. Calculate all side lengthes of the triangle ABF (marked red)

2. Calculate all side lengthes in triangle ABC (marked blue)

3. Calculate the side length FC in triangle ACF.

3. I don't understand how to get angels in ABC . I know how to find <C=135, but what about <A and <B?
And THANK YOU!!!

4. Originally Posted by Romanka
I don't understand how to get angels in ABC . I know how to find <C=135, but what about <A and <B?
And THANK YOU!!!
Complete my sketch by adding the given angles. Then you can calculate the interior angles in the two triangles in question.

5. Sorry, I'm completely stupid , but
how to get the angles <A =20 and <B=25? It can be easily proved that their sum is 45 degrees... Why aren't they 22 and 23 degrees, or 18 and 27, for example?
If I could catch about angles then I can find everything for the problem.
Thank you!

6. Originally Posted by Romanka
Sorry, I'm completely stupid , but
how to get the angles <A =20 and <B=25? It can be easily proved that their sum is 45 degrees... Why aren't they 22 and 23 degrees, or 18 and 27, for example?
If I could catch about angles then I can find everything for the problem.
Thank you!
According to your question the line AB and the North direction include an angle of 45° = NE.

According to your question the line AC and the North direction include an angle of 25°.

Therefore the angle at A is 45° - 25° = 20°

You'll get the interior angle at B in a similar way:

The line BC and South include an angle of 70°.
The line BA and South include anangle of 45°.

Therefore the $\displaystyle \angle(CBA) = 70^\circ - 45^\circ = 25^\circ$

etc + andsoon

7. Originally Posted by earboth
According to your question the line AB and the North direction include an angle of 45° = NE.

According to your question the line AC and the North direction include an angle of 25°.

etc + andsoon
that's exactly what I missed!
Thank you so much!