A private plane leaves the airport and flies 1.3 hours at 110 mph on a bearing if 40 degrees. Then it turns and continues another 1.5 hours at the same speed, but in a bearing if 130 degrees. Through what angle did the plane have to turn to head back to the airport?

Heading = 269.09 degrees. Plane will turn through an arc of 139.09 degrees. First, lets calculate the length of the 2 legs we know of. Leg1 = 1.3 * 110 = 143 miles. Leg2 = 1.5 * 110 = 165 miles. These distances are short enough that we can ignore the curvature of the earth and stick with euclidean geometry. Let's convert the polar coordinate of (40, 143) to Cartesian. (143sin (40), 143cos (40)) = (91.92, 109.54) So we know the plane flew 109.54 miles to the north, and 91.92 miles east. Lets see about the second leg. (165sin (130), 165cos (130)) = (126.40, - 106.06) So the plane then traveled 126.40 miles to the east, and 106.06 miles to the south. Adding those distances together gives (91.92 + 126.40, 109.54 - 106.06) = (218.32, 3.48) So the plane is now 218.32 miles to the east and 3.48 miles to the north of where it originally started. Let's calculate the angle needed to return by first calculating the tangent. 3.48 / 218.32 = 0.01594 atan (0.01594) = 0.91 degrees. Since the plane will have to be going almost due west, that means that the heading will be close to 270 degrees. And since it also needs to go slightly south, we'll subtract the 0.91 degrees from the 270 to get a heading of 269.09 degrees. In order to get to that new heading from it's current heading of 130 degrees, we simply subtract. So 269.09 - 130 = 139.09 degrees

Refer to the diagram shown below.

The first distance traveled is

a = (110 mi/h) * (1.3 h) = 143 mi

The second distance traveled is

b = (110 mi/h) * (1.5 h) = 165 mi

From geometry,

x = 40°

y = 180° - 130° = 50°

x + y = 40° + 50° = 90°

Therefore we have a right triangle.

By definition

tan θ = 165/143 = 1.1538

θ = tan⁻¹ 1.1538 ≈ 49°

Because the sum of angles in a triangle is 180°, therefore

z = 180° - (49° + 90°) = 41°

The angle through which the plane has to turn to head back to the airport is

φ = 360° - (y + z)

= 360° - (50° + 41) °

= 269°

Answer: 269°