I made a diagram of a parallogram, then drew in the diagonal to show the resultant, called it R
Looking at a triange, I have sides 20 and 500 with a contained angle of 105°
R^2 = 20^2 + 500^2 - 2(20)(500)cos105°
R = 505.546 mph
repeating for the side at 30
R2 = 508.59 mph
for first case: R = 505.546
sinØ/500 = sin105/505.546
Ø = 72.8°
so bearing = 72.8 - 30 = 42.8°
for 2nd case R2 = 508.59
sinØ/500 = sin105/508.59
Ø = 71.7
so bearing = 71.7 - 30
= 41.7
his bearing should be between 41.7° and 42.8°
A pilot is flying to a destination 700 miles away at a bearing of 105 degrees. The cruising speed of the plane is 500 mile per hour and the the wind is blowing between 20 and 30 miles per hour at a bearing of 30 degrees. At what bearing should the pilot aim the plan to compensate for the wind?
1 answer
1 answer