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3 years ago

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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?

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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