Answer to Question #346619 in Trigonometry for ariel

Question #346619

A plane leaves an airport with a bearing of S41W. After traveling 24 miles, the plane turns and travels on a bearing of N49W for 32 miles. At that time, what's the bearing of the plane from the airport?


1
Expert's answer
2022-06-01T02:54:30-0400

"\\angle B=41\\degree+49\\degree=90\\degree," so we have the right triangle ABC.

Let's find AC.

"AC=\\sqrt{AB^2+CB^2}=\\sqrt{24^2+32^2}=\\sqrt{1600}=40."

Now we can calculate "\\angle A:"

"sinA=\\frac{BC}{AC}=\\frac{32}{40}=0.8,"

"\\angle A \\approx53\\degree."




So "\\theta=180\\degree-(53\\degree+41\\degree)=86\\degree."

 The bearing of the plane from the airport is N86W.


Answer: N86W.



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