An airplane is headed due south with an airspeed of 300 kph. A wind is blowing from the west at 120 kph. What is the ground speed and direction of the plane?

323 kph with a directional bearing of S 22° E
328 kph with a directional bearing of S 68° E
420 kph with a directional bearing of S 22° E
420 kph with a directional bearing of S 68° E

a) 323 kph with a directional bearing of S 22° E

Step-by-step explanation:

to find the ground speed:

just do the pythagorean theorem.

c²=300²+120²

c²=104400

c=323

to find the directional bearing:

use the law of sines!

sin(90)/323 = sin(θ)/120

(*explanation: since the plane is heading due south and the wind is from due west, it makes a 90 degree angle; and the side opposite of the 90 degree angle is 323. set it equal to sin theta over 120; theta is the angle u r trying to find, and the side opposite of it is 120)

cross multiply:

120sin90=323sin(θ)

divide 323 on both sides:

sin(θ)=0.3715

use inverse sin to find out degree measure:

sin⁻¹(0.3715)=21.8

rounds up to 22.

A

Step-by-step explanation:

the guy above said so.