Please someone help me with question a) and b). I’ve been having so much trouble with these types of questions. The booklet says that the answers are a) 134degrees and b) 14.3km. I just don’t get how to get there. Thank you!

The bearing to take is 134 degrees and the shortest distance is 14.3 km
The attached diagram represents a labelled version of the ship movement
The angle at Q is
Q = 180 - 155 + 90 + 120 - 90
This gives
Q = 145
The distance PR is calculated as:
PR^2 =PQ^2 + QR^2 - 2 * PQ * QR * cos(Q)
This gives
PR^2 = 9^2 + 6^2 - 2 * 9 * 6 * cos(145)
Evaluate
PR^2 = 205.46
Take the square root
PR = 14.3 km
The angle at P is then calculated as:
QR/sin(P) = PR/sin(Q)
This gives
6/sin(P) = 14.3/sin(145)
Evaluate the quotient
6/sin(P) = 24.93
Make sin(P) the subject
sin(P) = 6/24.93
Divide
sin(P) = 0.2407
Take the arc sin of both sides
P = 14
The bearing is then calculated as:
Bearing = P + 120
This gives
Bearing = 14 + 120
Evaluate
Bearing = 134
Hence, the bearing to take is 134 degrees
In (a), we have
PR = 14.3 km
This represents the shortest distance
Hence, the shortest distance is 14.3 km
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