Question 1168157: Two ships leave a port O.
One ship travels on a bearing of 340° to a point P which is 50 km from O.
The other ship travels on a bearing of 060° to a point Q, 85 km from O.
(a) draw a diagram to represent the position of the port and the two ships. On your diagram carefully label north, the given angles and the distance travelled.
(b) calculate the distance PQ in km.
(c) Determine the bearing of P from O.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Draw this
Law of cosines with a being OP and b being OQ, c is PQ.
c^2=a^2+b^2-2abcosC
=50^2+85^2=2(4250)cos 80
=2500+7225-1476
=8248.99=8249
c=sqrt(8249)
=90.82 or 91 km.
law of sines for angle Q
sin 80/90.82=sin x/50
sin x=0.5421
take arc sin of that
x=32.83 deg
It can be shown on the drawing that the triangle in the second quadrant has angles 20, 67.17, and therefore 92.83 deg
That means the part of the triangle between the horizontal line at the y-axis to the bearing has to be 2.83 degrees.
The bearing from point Q (I think you meant that and not point O, which would simply bear 340, since it is given) is 270 deg + that 2.83 deg or 272.83 deg.
Alternatively, Point Q is 42.5 km N of point O and point P is 50 sin 70 or 46.98 km north of point O. PQ is the hypotenuse of a triangle with vertical distance 4.5 miles.
sine of the angle is 4.5/90.82, and the angle can be shown to be 2.84 degrees.
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