A plane is flying at a speed of 330330 miles per hour on a b

A plane is flying at a speed of 330330 miles per hour on a bearing of N6565°E. Its ground speed is 390390 miles per hour and its true course, given by the direction angle of the ground speed vector, is 3535°. Find the speed, in miles per hour, and the direction angle, in degrees, of the wind.

A plane is flying at a speed of 330 miles per hour on a bearing of N65°E. Its ground speed is 390 miles per hour and its true course, given by the direction angle of the ground speed vector, is 35. Find the speed, in miles per hour, and the direction angle, in degrees, of the wind. The speed of the wind is miles per hour. (Do not round until the final answer. Then round to the nearest tenth as needed.) The direction angle of the wd is Do not round until the final answer. Then round to the nearest tenth as needed.)

Solution

velocity of plane with respect to groud, v_pg =(390cos35^o)i+(390sin35^o)j

velocity of plane with respect to wind, v_pw =(330cos(90-65)^o)i+(330sin(90-65)^o)j

v_pw =(330cos25^o)i+(330sin25^o)j

velcity of wind with respect to ground =v_wg

v_pw +v_wg=v_pw

=>(330cos25^o)i+(330sin25^o)j +v_wg=(390cos35^o)i+(390sin35^o)j

=>v_wg=(390cos35^o-330cos25^o)i+(390sin35^o-330sin25^o)j

speed of wind =|v_wg|

speed of wind =[(390cos35^o-330cos25^o)²+(390sin35^o-330sin25^o)²]

speed of wind =86.7 miles per hour

direction angle of wind =tan^-1[(390sin35^o-330sin25^o)_{/(390cos35^o-330cos25^o)}]

direction angle of wind=76.4^o