Find the speed and direction of the air currents in mi/hr due north

Question

Use your knowledge of bearing, heading, and true course to sketch a diagram that will help you solve the problem.

A plane has an airspeed of 195 miles per hour and a heading of 22.0°. The ground speed of the plane is 221 miles per hour, and its true course is in the direction of 40.0°. Find the speed and direction of the air currents, assuming they are constants. (Round your answers to one decimal place.)

 ______mi/hr at  _______ ° from due north

Answer 1

The angle between the directions is 40° - 22° = 18°.

Then, we have two sides, 195 and 221, and the included angle, so we have to use the cosine law to get the opposite side.

a = √(b²+c²-2bc cos A) = √(195²+221²-2*195*221 * cos 18°) = 69.9602662982405

This is the windspeed.

Then, to get the direction, we need to get one of the other angles. As the angle opposite 221 might be obtuse (it turns out that it is), we will use the angle opposite 195.

sin 18°/69.9602662982405 = sin B/ 195.

B = sin-1(195 sin 18°/69.9602662982405) = 59.4653384196936 (note that 18+60=78 < 90 so the other angle is obtuse as stated).

Then, as this is the internal angle, the direction of the wind will be 180° + (40° - (180° - 59.4653384196936)) = 40° + 59.4653384196936 = 99.4653384196936°

(Note that the internal angle between the 22° angle and the 99.4653384196936° angle is 180° + 22° - 99.4653384196936° = 102.534661580306, and 102.534661580306 + 18 + 59.4653384196936 = 180.

windspeed 69.9602662982405

wind direction 99.4653384196936°