Question

A plane takes off and travels at an angle of 40° north of east at 110mph for 2 hours. It then adjusts its path to head 10° west of north and travels in that direction for half an hour at a speed of 100mph. Approximately how far away is the plane from its starting point?

Answer 1

Answer: 248.80 miles ( Approx)

Explanation:

Let A shows the initial position of the plane, B shows the position of plan after traveling at an angle of 40° north of east at 110 mph for 2 hours and C represents the final position of plan.

Then According to the question,

`AB = 220` miles ( because plane traveled for 2 hours with the speed of 110 mph)

`BC = 50` miles ( Because plane traveled for 1/2 hour with the speed of 50 mph)

Now,
`m\angle ABC = 40^(\circ)+80^(\circ)=120^(\circ)` ( shown on diagram)

Thus, By cosine law,

`AC^2 = AB^2 + BC^2 - 2AB. BC cos (120^(\circ) )`

`AC^2 = 220^2 +50^2 - 2* 220* 50* cos (120^(\circ) )`

`AC^2 = 61900`

`AC=248.797106092 \approx 248.80`

Thus, the plane is 248.80 miles from its starting point.