Question

A pilot wants to fly on a bearing of 74.9 degrees. By flying due east, he find s that a 42-mph wind, blowin from the south, puts him on course. Find the airspeed and the groundspeed.

Answer 1

The bearing shows the angle from north to the line (c), you want to see the angle inside the triangle so that will be 90-[bearing]. 

a=42 (42 mph wind blowing north) 

A= 90-74.9 
A= 15.1 degrees 

Ground speed is the speed the plane is going including the wind. 
ground speed = c 
Airspeed = b 

You have the angle, and you have the "Opposite" and want to find the "Hypotenuse". 
SOH CAH TOA 

Sin x = Opp/Hypot 

Sin(15.1) = 42/c 
c = 42/Sin(15.1) 
c = 161.53 

Ground speed of the plane is 161.53 mph 

Airspeed = b 

Tan x = Opp/Adj 
Tan (15.1) = 42/b 
b = 42/Tan(15.1) 
b = 161.53 

Airspeed = 161.53 mph 

(so the answer is ground speed of 161.53mph)