A pilot flies in a straight path for 1 h 30 min She then mak

A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 11° to the right of her original course, and flies 2 h in the new direction. If she maintains a constant speed of 575 mi/h, how far is she from her starting position? (Round your answer to the nearest mile.)

Solution

Lets asssume a triangle with leg x = 1.5* 575 = 862.5 miles

2nd leg of triangle y = 2*575 = 1150 miles

Triangle is xyz where we have to find side z using cosine rule:

z is the distance from the starting point

Angle z = 180 -11 = 169deg

z^2 = (1.5*575)^2 + (0.5*575)^2 - 2*(1.5*575)*(0.5*575)cos(169)

z =2003.42 miles = 2003 miles