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October 21, 2014

October 21, 2014

Posted by **p** on Sunday, September 15, 2013 at 8:01am.

150 km.

a How far is the plane from its starting point?

b What direction is the plane from its starting point?

- trg -
**Reiny**, Sunday, September 15, 2013 at 8:59amI will convert the directions to vectors using conventional angles

first leg: (80cos20° , 80sin20°)

2nd leg: (150cos260° , 150sin260°)

resultant

= (80cos20° , 80sin20°) + (150cos260° , 150sin260°)

= (49.128, -120.3596)

magnitude = √(49.128^2 + (-120.3596)^2) = 130 km

direction:

tanØ = -120.3596/49.128

Ø = -67.8° or 292.2°

or

S 22.2° E

- trg -
**Reiny**, Sunday, September 15, 2013 at 9:09amor

Make a sketch , let the original position be A, end of first leg B, and C the final position.

I have a triangle ABC with angle B = 60°

AB = 80 and BC = 150

by the cosine law:

AC^2 = 80^2 + 150^2 - 2(80)(150)cos60

= 16900

AC = √16900 = 130

by sine law:

sinA/150 = sin60/130

sinA = 150sin60/130 = .99926

angle A = 87.8°

looking at your diagram, subtracting 20° yields the same angle of 67.8° as in my first solution.

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