Question 357331
A plane needs to fly to a town located 200 km away in the direction 50 degrees east of north. There is westerly wind blowing at 60 km/h. 
Determine the plane's airspeed and the direction in the degrees east of north, so it can fly directly to the town in 1/2 hour. 
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The groundspeed has to be 400 km/hr (200 km/0.45 hr)
The wind is 60 from the west.
The angle between the flight path and the wind direction is 40º where they meet at the town.
This makes a triangle with sides meeting at the town of 400 and 60 with the angle between them of 40º.
Use the Cosine Law to find the 3rd side, with is the plane's airspeed.
{{{a^2 = 60^2 + 400^2 - 2*60*400*cos(40)}}}
{{{a^2 = 126830}}}
airspeed = 356.13 km/hr
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Use the Law of Sines to find the heading of the plane:
356.13/sin(40) = 60/sin(B)
sin(B) = 60*sin(40)/356.13 = 0.10829
B = 6.217º
Heading = 50 - B = 43.8º
or 43.8 East of North
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PS  Wind info is always given as the direction it's from.
A "westerly wind" of 10 knots would be given as "winds 270 at 10"
West is 270, and 10 is the windspeed in knots.
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