The bearing of Q from P is 300 degrees and the bearing of R from Q is 120 degrees, if Q is equidistance from P and R, find the bearing of R from P.
2 answers
Following your instructions had me placing R at P.
Let: PQ = 1[300o], QR = 1[120o].
PR = 1[300] + 1[120].
PR = (1*sin300+1*sin120) + (1*cos300+1*cos120)i
PR = 0 + 0i = 0.
The vectors are 180o out of phase and equal magnitude.
Bearing and magnitude = 0.
PR = 1[300] + 1[120].
PR = (1*sin300+1*sin120) + (1*cos300+1*cos120)i
PR = 0 + 0i = 0.
The vectors are 180o out of phase and equal magnitude.
Bearing and magnitude = 0.
11 answers