A plane flies 430 miles at a bearing of N27°E, then turns and flies N63°E for 135 miles. Find the distance and the bearing from the starting point to the ending point ?

Question

I found distance to be 450.69 mi by using A^2+B^2=C^2 because it is a right triangle. I am confused where to place the bearing angle therefore how to solve for the bearing

Answer 1

these are headings, not bearings.

I do not get a right triangle. I get an angle of 27+90+27= 144 at the turning point.

To get the final location, convert each side to its x- and y-components.

Add them up, and you can then get the angle θ using tanθ = y/x

and then the bearing is N(90-θ)E as usual