Answer: 248.80 miles ( Approx)
Step-by-step 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,
[tex]AB = 220[/tex] miles ( because plane traveled for 2 hours with the speed of 110 mph)
[tex]BC = 50[/tex] miles ( Because plane traveled for 1/2 hour with the speed of 50 mph)
Now, [tex]m\angle ABC = 40^{\circ}+80^{\circ}=120^{\circ}[/tex] ( shown on diagram)
Thus, By cosine law,
[tex]AC^2 = AB^2 + BC^2 - 2AB. BC cos (120^{\circ} )[/tex]
[tex]AC^2 = 220^2 +50^2 - 2\times 220\times 50\times cos (120^{\circ} )[/tex]
[tex]AC^2 = 61900[/tex]
[tex]AC=248.797106092 \approx 248.80[/tex]
Thus, the plane is 248.80 miles from its starting point.
