Answer
The pilot turned to head back for the carrier at 3:21 pm
Work Step by Step
We can find the distance traveled by the carrier in 2.6 hours:
$(2.6~h)(32~mph) = 83.2~miles$
We can find the distance traveled by the plane in 2.6 hours:
$(2.6~h)(520~mph) = 1352~miles$
Let $x$ be the distance traveled by the plane on a bearing of 338. The plane flies from point A to point B, and then turns and meets the carrier at point C.
We can find the angle $\angle BAC$ on the diagram:
$\angle BAC = 30^{\circ}+(360^{\circ}-338^{\circ}) = 52^{\circ}$
We can use the law of cosines to find the value of $x$:
$(1352-x)^2 = x^2+(83.2)^2-(2)(x)(83.2)~cos~52^{\circ}$
$1,827,904-2704~x+x^2 = x^2+6922.24-102.446~x$
$2601.554~x = 1,820,981.76$
$x = \frac{1,820,981.76}{2601.554}$
$x = 700~miles$
We can find the time it takes the plane to fly 700 miles:
$t = \frac{700~mi}{520~mph} = 1.346~hours = 1~hr~21~min$
Since the pilot left at 2 pm, the pilot turned to head back for the carrier at 3:21 pm
