A pilot diverted a fighter jet 20.0 km [S 38.0° E] to avoid thunderstorm, and then flew 60.0 km [S 55.0° W]
to get back on track. If the diversion lasted a total of 2.00 minutes, what was the average velocity of the
jet during the diversion? Show all your work
DRAW IT !!!!
A at origin
B at 20 km 38 deg E ofS
C at 60 km 55 deg W of S from B
AC is original path
angle B = 180 - 38 - 55 = 87 deg
law of cosines
AC^2 = 20^2 + 60*2 - 2*20*60 cos 87
find AC (almost a right triangle hypotenuse for check, about 63 km)
so average speed =63/2 km/(1/3 hr) = 94.4 km/hr
get direction from the triangle law of sines