Asked by Simon Silas Bako

A ship leaves port and travels 21km on a bearing of o32 degrees and then 45km on a bearing of 287 degrees.
a. calculate its distance from the port.
b. calculate the bearing of the port from the ship
Answered by oobleck
ships travel on headings, not bearings.

The angle turned is 75°, so the distance d from port is
d^2 = 21^2 + 45^2 - 2*21*45*cos75° = 1977
so, d = 44.46 km

21 @ 32° = <21 sin32°, 21cos32°>
45 @ 287° = <-75cos17°, 75sin17°>
add the x- and y-displacements, to get the final location. Then you can calculate the bearing (back whence it started)
Answered by Henry2,
a. d = 21[32] + 45[287].
X = 21*sin32 + 45*sin287 = -31.9km.
Y = 21*Cos32 + 45*Cos287 = 30.97km.
d = sqrt(X^2 + Y^2) = 44.46 km.

b. Tan A = X/Y
A = -45.8o = 314.2o CW from +y axis.













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