Question

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?

Answer 1

$\angle B=41\degree+49\degree=90\degree,$ 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.