precalculus
posted by amaa .
A pilot flies in a straight path for 2 h 30 min. She then makes a course correction, heading 7 degrees to the right of her original course, and flies 3h in the new direction. If she maintains a constant speed of 695 mi/h, how far is she from her starting position?

let AB be the original path, then veer off at 7° for BC.
Join AC
I see a long skinny triangle ABC with angle ABC = 173°
The lengths of AB and BC are 2.5(695) and 3(695) respectively, but we can just ignore the 695 for the time being and use the smaller similar triangle to keep the numbers low.
by cosine law
AC^2 = 2.5^2 + 3^2  2(2.5)(3)cos173
= 15.25  (14.88819...)
= 30.13819..
AC = 5.489826..
so the distance is 5.489..(695) = 3815.4 miles