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.

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: