A plane flies 1.3 hours at 110 mph on a course of 38°. It then turns and flies 1.5 hours at the same speed on a course of 128°. How far is the plane from the starting point?

Answer 1

At first the plane was flying on a course of 38° with the speed of 110 mph for 1.3 hr. It covered distance in this direction (d = v·t) d_AC = 110 x 1.3 = 143 (miles). Similarly we could find that on a course 128° the plane covered d_CB = 165 miles. Now we have a triangle ACB. In this triangle the length of side AC is 143 miles, length of side CB is 165 miles and angle ACB has a measure of 104° (180 - 38 - 38 = 104). In an obtuse triangle ACB we know the length of two sides and the value of angle between them. Hence we can use the cosine theorem in the form c² = a² + b² -2a·b·cos Θ. to find the length of the side AB.

Here a = 143, b = 165 and Θ = 104. If you plug in the values, you obtain

c² = (143)² + (165)² - 2·143·165·cos (104°)

Finally, c = 243 miles.

Answer: the plane is 243 miles from starting point