It is adding vectors
150@someangle+ 25@(340-180)=something@70
so break these up into N, E components.
Measure angles from true NORTH.
North equation
150cosTheta+25cos(340-180)=N?*cos70
Solve for N?
East equation
150sinTheta+25sin(340-180)=E?*sin70
Finally, the third equation:
N?/E?=tanTheta
Three equations, three unknowns. Lord, it looks like fun, doesn't it?
Graphically, this is a piece of cake.
Draw the wind vector to the origin. Draw the resultant line at 70 degrees.
Now, use a compass to make an arc of 150 to find where it intersects the 70 degree ray, if the compass arc originates from the head of the wind.
Method three.
You know two sides, one angle (between wind and resulatant) SAS triangle.
Law of cosines:
150^2=H^2+25^2-2H*25*cos(340-180-70)
and you solve H from that, H is the magnitude of the speed over ground.
An airplane has an airspeed of 150 km/hr. It is to make a flight in a direction of 70 degrees while there is a 25 km/hr wind from 340 degrees. What will the airplane's actual heading be?
Please Explain this Problem to me!!!!
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