P, Q and R are points on the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from P is 060°. If /PQ/=5m and /QR/=3m

a.) Sketch the diagram
b.) Find the bearing of R from P, correct to the nearest degree.

11 answers

Typo, impossible as stated. If the bearing of R from P is 60 why do you ask for it in part b?

b. PR = PQ + QR = 5m[150o] + 3m[60o],
PR = (5*sin150+3*sin60) + (5*Cos150+3*Cos60)I = 5.1 - 2.83i.

Tan A = X/Y = 5.1/-2.83,
A = -61o = 61o E. of S. = 119o CW(bearing).

NOTE: Where you typed "bearing of R from P" is an error; it should be
"bearing of R from Q is 060."

Bearing

P,Q and R are point in the same horizontal plane the bearing of Q from P 15° and the bearing of R from Q is 160° if PQ = 5km and QR = 3k find the bearing of R from Q

Don't understand

P,Q and R are point in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 60° If /PQ/=5m and /QR/=3m. Find the bearing of R from P, correct to the nearest degree

None

calculate it

No idea

To mathematics

Tan¤=3/5
tan¤=0.6m
¤=tan+0.6m
¤=30.9degree=31degree
bearing of R from P
=180-(30+31)
180-61=119degree
therefore bearing of R from P=119degree