Tough Trigonometry Word Problem - Please help?

Question

I need some ideas on how to go about solving this problem, any help at all will be greatly appreciated, thanks!

A plane has airspeed of 210 mph and a heading of 310 degrees. The wind is blowing at a constant 15 mph in the direction of (North 35 degrees East). Find the ground speed and true course of the plane.

Answer 1

Nothing tough here, except your odd way of describing the wind direction.
The plane is going east at 210(cos310) and south at 210(sin310) (south because it's negative. Decompose the wind speed in the same way, add the east-west speeds and north-south speeds, and use Pythagoras and tangent to find speed and direction.

Answer 2

Draw a line and call it side A. It represents the vector 310 degrees at 210 mph. From the top of this line draw another line (B) to the right and make it shorter. This represents the wind vector 35 degrees and 15 mph. Ok. There are 180 degrees along the line A which is pointing to 310. Due north is 360 so there are 50 degrees to north and the wind (B) is another 35 degrees off that. Then the angle between the A vector and the B vector is 50+35 or 85 degrees. That means the complementary angle on the other side of B is 180 - 85, or 95 degrees. The line "C" that connects the bottom of A to the right end of B represents the ground speed of the plane. You now have a triangle, two sides are known as is one angle. I hope you can take it from here.

Answer 3

consider that the azimuth is based in the + x-axis, counter clockwise direction
form a vector addition
form a triangle where the sides are 210 and 15 with the angle of 85 deg in between them
by Cosine law
speed^2 = 210^2 + 15^2 -2 (210) (15) cosine 85 = 44874.08

speed = 211.84 mph

solving course direction. by sine law
sin A = 210 ( sin 85 / 211.84 ) = 0.9875419
A = 80 deg 57'
angle = 90 - 35 deg - 80 deg 57' = 25 deg 57'

direction of airplane : South 25 deg 57' East

Answer 4

Draw axes for N,S,E and W (centre 0)
OP in 4th quadrant (SOE quadrant)
Angle POS = 40°
Draw BP a vertical line thro` P (B above P)

Angle OPB = 40°
Angle BPA = 35° (from N 35°E info.)
A in 4th quadrant to right of BP
Consider triangle OPA:-
Angle OPA = 75°
OP = 210
PA = 15
Using cosine rule:-
OA² = 210² + 15² - 2 x 210 x 15 x cos 75°
OA = 206.7 mph (ground speed)

15 / sin POA = 206.7 / sin 75°
sin POA = 15 sin 75° / 206.7°
POA = 4°
Course of plane = 270° + 40° + 4°
Course of plane = 314°