SOLUTION: A plane is flying west at 200 mph. The wind begins blowing S30°W at 25 mph. What is the ground speed of the plane now? (nearest tenth) When you solve the triangle, wh

The black vector is the plane's (heading) velocity vector of 200 mph due West.
The blue vector is the wind's vector of 25 mph.
The red vector is the resultant vector (plane's actual velocity vector).

The pilot is trying to fly in the direction of the black vector, but the wind 
is blowing him in the direction of the red vector.

S30oW means a direction which "swings" from due South (straight down)
towards due West (to the left).

So the angle between the green line (pointing south) and the blue (wind) vector
is 30o.

So the angle between the blue vector and the black vector is
90o more or 120o.

So we have side-angle-side and so we use the law of cosines:



That comes out to 213.6 mph for the ground speed.  So the wind is speeding
him up but blowing him toward the south slightly.

Can you work out the angle swinging from due South left toward due West,
that the red vector is pointing in?  

Hint: Draw a vertical line from the right vertex straight downward due South. 
Then use the law of sines to find the angle between the line you drew
and the red vector. That will be the angle to put between S and W.  You will
need to first find the smallest angle in the triangle between the black and red
vectors.

I don't have time. 

Edwin