Posted by Physics on Friday, September 18, 2009 at 9:18pm.
I gues I'm looking at this the wrong way i do not understand the work you do to get to the answer
Question:
Find the air velocity of a plane that must have a realtive ground velocity of 250.0 h^1 km norht if it encounters a wind pushing it toward the northeast at 75 h^1 km
The Work:
I do not follow it one bit...
Va = Vg  Vw
= 250 h^1 km north  75.0 h^1 km northeast
= 250.0 h^1 km north + 75.0 h^1 km southwest
Va^2 = Vg^2 + Vw^2  2VgVw cos A
= (250.0 h^1 km)^2 + (75.0 h^1 km)^2  2(250.0 h^1 km)(75.0 h^1 km)(cos 45.0)
= 62 500 h^2 km^2 + 5 625 h^2 km^2 + 26 516 h^2 km62
= 41 609 h^2 km^2
Va = 204 h^1 km
Vw/sin W = Va sin A
sin W = Vw sin A/Va
= (75 h^1 km)(sin 45.0)/(204 h^1 km)
= (75 h^1 km)(.707)/(204 h^1 km)
= 0.260
B = 15.1
Answer = 204 h^1 km, 15.1 degrees west of north
ok i was lost on line 2 and three were the added and subtracted 75 with two different directions why are there two
line 3 were did this come from
 2VgVw cos A
thanks!

Physics  MathMate, Friday, September 18, 2009 at 9:35pm
Pythagoras theorem applies to the sides adjacent to the right angle:
a²+b²=c²
The extension when the angle is not a right angle is called the cosine rule, which states that:
a²+b²2ab cos(C) = c²
which is what line three is doing, the angle between NE and N being 45°.
The resultant (air velocity) was calculated at 204 km/hr.
The line
Vw/sin W = Va sin A
should read
Vw/sin W = Va/sin A
which is the application of the sine rule to find the angle of a triangle in which one side and the opposite angle is known, as well as the side opposite to the angle sought.
I have not checked the numerical calculations. You can do the checking as an exercise.
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