Question

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?

Answer 1

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