Apilot flies in a straight path for 1 h 30 min. she then makes a course correction, heading 10 degrees to the right of her original course, and flies 2 h in the new direction. if she maintains a constant speed of 600 mi/h, how far is she from her starting position?

2,092.17 miles

Explanation:

To solve this easily, let's do a graph about it. The picture attached shows the travel of the pilot.

Now, let's calculate the distance from A to B (t = 1h 30 min or 1.5 h)

d1 = 600 * 1.5 = 900 mi.

The distance from B to C:

d2 =600 * 2 = 1200 mi.

Now, we need to know the distance fro A to C. This is the part of the triangle that must be calculated. To do this, we can use several theorems, but with the cosen law we can solve this easily. The cosen law is the following:

d3² = d1² + d2² - 2*d1*d2cosB

B is the angle formed in the triangle. we know that the plane do an inclination of 10° about the horizontal, therefore, the remaining degrees will be B, thus:

B = 180 - 10 = 170°

Remember that we used 180° because the inner angles of a triangle always have to sum 180°.

Now that we have this, let's calculate the distance 3, which will be the distance from the starting point:

d3² = 1200² + 900² - 2 * (1200) * (900) cos170

d3² =2,250,000 + 2,127,184.75

d3 = √4,377,184.75

d3 = 2,092.17 mi

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y