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 a

Question

2 years ago

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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 a

irport 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.

Mathematics

1 answer:

Karolina [17] · Answer 1 · 2021-11-30T03:07:32+03:00

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