From a point X, a boat sails 6km on a bearing of 037° to a point Y. If it then sails 7km from Y on a bearing of 068° to a point Z.
Calculate the;
1.distance XZ correct to two dercimal places
2.the bearing of Z from X to the nearest degree
As always, draw a diagram. In triangle XYZ, it should be clear that angle Y = 149°
So, using the law of cosines,
XZ^2 = 6^2 + 7^2 = 2*6*7 cos149°
If we consider X as being at (0,0), then Z is at
(6cos53°+7cos22° , 6sin53°+7sin22°) = ( 10.1, 7.4)
So the bearing of Z from X is 90-θ, where tanθ = 7.4/10.2
All angles are measured CW from +y-axis.
1. XZ = XY+YZ = 6[37]+7[68o]
XZ = (6*sin37+7*sin68)+(6*cos37+7*cos68)i
XZ = 10.10+7.41i = 156.98km[54o].
2. Bearing(direction) = 54 deg.