Find Distance and Bearing from City A to C | Plane Navigation Problem

[Phyzwizz]

Problem
Distance:
From city A to city B, a plane flies 650 miles at a bearing of N 48° E. From city B to city C, the plane flies 810 miles at a bearing of S 65° E. Find the distance from A to C and the bearing from A to C.

Work so far
So far I've gotten the distance from A to C already, but I'm having trouble figuring out what to do, to find the bearing. Its sort of difficult to show what I did to get the sides because it involves drawing a series of other triangles and finding the lengths of some of their sides and adding them in some places and subtracting in others. But the tactic I took in solving for the length was setting A to (0,0) and attempting to find the x and y coordinates for C separately.

What I got for the distance from A to C is, AC≈ 1220.671.

Thanks, I hope someone can help me.

 

[Dick]

What did you get for the coordinates of C? Just draw one right triangle with hypotenuse AC.

 

[verty]

The bearing is just the angle at A between North and AC. In the right angled triangle with AC the hypotenuse, you know two lengths, so you can use trig ratios to find the angle.

 

[Phyzwizz]

The coordinates I got for C are...

X - ≈1217.153 came from (650cos 42°)+(810cos 25°)

Y - ≈140.723 came from (650sin 42°)-(810sin 25°)

 

[Phyzwizz]

Oh, thanks Vertigo, I guess I just need to think about the problem a little more next time. That really is pretty simple.

I just do sin θ = opp/hyp and then I can switch it and make it arcsin opp/hyp= θ right?

θ=theta
opp= opposite
hyp= hypotenuse

 

[verty]

Oh, thanks Vertigo, I guess I just need to think about the problem a little more next time.

That's okay, I'm sure you'll find these questions quite easy from now on. And I'm sure you'll learn soon in class more about arcsin and quadrants and such.