A plane is flying with an airspeed of 190 miles per hour and heading 140°. The wind currents are running at 25 miles per hour at 165° clockwise from due north. Use vectors to find the true course and ground speed of the plane. (Round your answers to the nearest ten for the speed and to the nearest whole number for the angle.)

Question

Reiny · Answer

ground speed = (190cos140°, 190sin140°) + (25cos165°,25sin165°)
= (-169.7, 128.6)

magnitude = √( (-169.7)^2 + (128.6^2) ) = appr 212.9 mph
for direction:
tanØ = -128.6/169.7
angle Ø in standard position = 37.16°

or

after sketching the parallelogram and labelling the angles
|R|^2 = 190^2 + 25^2 - 2(25)(190)cos155°
= 45334.92..
R = √45334.92 = appr 212.9 km/h , just as above

The acute angle of the needed triangle, by the sine law:
sin x /25 = sin155/212.9
angle x = 2.84°
add that to 140° to get a heading of 142.84°

Notice from my vector calculation : 180-37.16 ° = 142.84°

oobleck · Answer

just add the vectors. I get 212.92 @ 142.84°