Answer:
Ground speed = 131.25 km/h
Bearing = 92.68º
Step-by-step explanation:
If we imagine a triangle to depict the question with one side 134 km/h at an angle of 78º to the right of vertical (the plane's speed vector).
At the end, we have a line 21 km/h at an angle of 180 (from north), a vertical down from the end of the plane's vector. This vertex is angle B
This makes a triangle with the interior angle B of 78º.
The 3rd side is the ground speed which we can find using the Cosine rule:
b² = a² + c² - 2•a•c•cos(78)
b² = 134² + 21² - 2•134•21•cos(78)
b² = 17226.873
b = 131.25 km/h
The true course bearing" is the ground track. Thus;
Using the Sine rule to find angle A, we have;
(sin(A))/34= sin(78)/b
Thus,
sin(A) = (34 x sin(78))/131.25
sin(A) = 0.2534
A = sin^(-1)0.2534
A = 14.68º
Ground track = A + 78 = 14.68 + 78 = 92.68º
