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solmaris [256]
3 years ago
12

A jet leaves a runway whose bearing is N 3232degrees°E from the control tower. After flying 77 ​miles, the jet turns 90degrees°

and flies on a bearing of S 5858degrees°E for 88 miles. At that​ time, what is the bearing of the jet from the control​ tower?
Physics
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

N(80.8^{\circ})E

Explanation:

Bearing from control tower= N(32^{\circ}+\theta})E

But tan \theta=\frac {opposite}{adjacent}

tan \theta=\frac {88}{77}

\theta=tan^{-1}(\frac {88}{77})=tan^{-1}(1.142857143)=48.81407483^{\circ}\approx 48.8^{\circ}

Therefore, the bearing from control tower=N(32^{\circ}+48.8^{\circ}})E

Bearing=N(80.8^{\circ})E

The direction is North East

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