The bearing of point Y from a point X is 048°. If Y is 115km away from point X and the bearing of point Z from Y is 120° at a distance of 75km. Find the distance between X and Z, using a scale of 1cm to represent 20km

Answer 1

Did you make your sketch?
It should show an angle of 108° between XY and YZ
by the cosine law:
XZ^2 = 115^2 + 75^2 - 2(115)(75)cos 108°
XZ = .....

or

vector XY = 115(cos42, sin42) = (85.46165... , 76.95...)
vector YZ = 75(cos-30 , sin -30) = (64.9519.. , -37.5)
XZ = (150.4133.. , 39.45..)
|XZ| = √(150.4133..^2 + 39.45..^2) = .....

You must get the same answer, I did.

Answer 2

D = 115km[48o] + 75km[120o],
X = 115*sin48 + 75*sin120 = 150.4 km,
Y = 115*Cos48 + 75*Cos120 = 39.45 km,
D = 150.4 + 39.45i = 155.5km[75.3o] = Distance from X to Z.

Tan A = X/Y For bearing notation.
A = 75.3o.