Check work, please. Vector Question?

Question

I have a question that reads: "A pilot wishes to maintain a ground speed of 425 mph and fly at a bearing of 155 degrees. The wind is blowing at 30 mph on a bearing of 45 degrees. Find the airspeed and bearing that the pilot should choose"

I made a vector diagram and using the law of cosines I foudn the magnitude to be 446.72 mph and the bearing to be 152.28 degrees.

If anyone could take a look at this problem and see if I've done this right I would really appreciate it!

Answer 1

Diagram
In Triangle OBA:-
OA = 425 /_155°
OB = 30/_ 45°
/_AOB = 110°
AB² = 425² + 30² - (2 x 30 x 425 cos 110°)
AB² = 180625 + 900 + 8722
AB = 436.2 mph (air speed)
30/ sin A = 436.2 / sin 110°
sin A = 30 sin 110° / 436.2
A = 3.7°
Bearing = 90° + 65°+ 3.7°
Bearing = 158.7°

Answer 2

Let
w = wind
a = airplane - airspeed and heading
r = resultant = airplane's actual ground speed and direction

r = a + w
a = r - w

Calculate the compass bearings for the airplane, North and East.

N = 425cos(155°) - 30cos(45°) = -406.3940129
E = 425sin(155°) - 30sin(45°) = 158.3995578

The magnitude of the airplane is:

| a | = √[(-406.3940129)² + (158.3995578)²]
| a | = 434.7837045 mph

The directions of the airplane is:

tanθ = E/N

θ = arctan(E/N)
θ = arctan(158.3995578 / -406.3940129) = 159.18°
___________

As a check, the airspeed should be greater than the groundspeed since the angle between the wind and the direction of the plane is greater than 90°.

Also, the resulting direction of the plane (155°) should be between the direction of the air (45°) and the heading of the plane (159.18°).

Answer 3

you draw a diagram illustrating with 1 triangle , the final vector x . use cos law to find the the magnitude of x
x^2=425^2+30^2-2(425)(30)cos110
x=436.2mph
use sin law to find the angel
sina/425=sin110/436.2
x=66.3 degree
66.3-(180-110-45)=41.3degree
180-41.3=138.7 degree
so the answer is 436.2 at 138.7 degree