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

`N(80.8^(\circ))E`

Step-by-step 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