Answer:
Therefore, the resultant ground speed of the plane after accounting for the wind is approximately 825.8 mph, and the resultant direction is 53.8° East of North.
Explanation:
Based results, it appears that the question is asking for the resultant ground speed (magnitude) and direction of a plane flying at 700 mph on a course heading 10° East of North, while experiencing a wind blowing in a direction of 72° West of North at a speed of 150 mph. Here is how to solve this problem:
Convert the course heading and wind direction angles to Cartesian coordinates using the standard unit vector notation i and j. The course heading is (cos(10°)i + sin(10°)j) and the wind is (-cos(72°)i + sin(72°)j).
Add these vectors together to get the resultant vector. The sum is (-478.3i + 684.2j).
Find the magnitude of the resultant vector using the Pythagorean theorem. The magnitude is approximately 825.8 mph.
Find the angle that the resultant vector makes with respect to the North using the inverse tangent function. The angle is approximately 53.8° East of North.
Therefore, the resultant ground speed of the plane after accounting for the wind is approximately 825.8 mph, and the resultant direction is 53.8° East of North.
( Sorry if the answer is wrong ).