Question

A plane leaves airport A and travels 570 miles to airport B on a bearing of Upper N 32 degrees Upper E. The plane later leaves airport B and travels to airport C 390 miles away on a bearing of Upper S 73 degrees Upper E. Find the distance from airport A to airport C to the nearest tenth of a mile.

Answer 1

Answer:

769.46 miles

Step-by-step explanation:

Please have a look at the photo attached in this answer

The missing part of the angle is 32 degrees, so the total angle of B is:

32 + 73  = 105 degrees.

Let x is the distance from airport A to airport C to the nearest tenth of a mile, we use cosine law to find x:

$x^{2}$ = $570^{2} + 390^{2}$  - 2(570)(390)cos(105)

<=> $x^{2}$ = 592070

<=> x = $\sqrt{592070}$ = 769.46 miles