A plane has an airspeed of 134 km divided by h. It is flying on a bearing of 78 degrees while there is a 21 km divided by h wind out of the northeast (bearing 225 degrees). What are the ground speed and the bearing of the plane?

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

probably the last option

step-by-step explanation: