Apilot 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.

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)

Use an app called “PhotoMath” just take a picture of it and it’ll show you the answer. :)