Physics Help Please!!!!!!!!??????????

Please read :

These problems can be very confusing.

Velocity of plane relative to ground = vpg
Velocity of plane relative to air = vpa
Velocity of air relative to ground = vag

vpg =vpa + vag (note the pattern of letters)
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When drawing triangle ABC let:
vpa = AB
vag = BC
then that means vectorAB + vectorBC = vectorAC. So we can say
vpg = AC
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vpa = 200km/h south
With a scale of 1mm representing 1km/h, vpa is line AB, 200mm long pointing south.
A
.
.
B

vpg = 125km/h 15.0∘ east of south.
Represent vpg as line AC, 125mm long pointing 15.0° east of south(hour-hand at 6:30 on a 12-hour clock face).
A

. C

The complete triangle is therefore is like this:
A

. C
B
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We want vag (BC). Using the cosine rule:
BC² = AC² + AB² - 2.AC.AB.cos(∠A)
. . . .= 125² + 200² = 2x125x200cos(15°)
. . . .= 7329
BC = 85.6km/h.
This is the answer to part A. You might be expected to round to 86km/h.
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Using the sine rule:

AC/sinB = BC/sinA

125/sin B = 85.6/sin(15°)
sinB = 0.378
B = sin⁻¹ (0.378) = 22° approx,

So (referring to the direction of BC on the triangle-diagram above) we see the wind direction is 22° east of north.