in triangle QPR, angle QPR is 75+50 = 125°
All we have so far is one side (PQ=90) and one angle.
That does not determine a triangle.
Figure out what is missing, then use the law of sines or law of cosines as needed.
Three towns pqr are such that the distance between p and q is 90km.if the bearing of q is 075 and the bearing of R from P is 310 find the
A.distance between Q and R
B. Bearing of R from Q
2 answers
A. QR/sin P = PQ/sin R.
QR/sin125 = 90/sin40
QR = 115 km.
B. PR/sin Q = PQ/sin R.
PR/sin15 = 90/sin40
PR = 36 km.
QR = QP+PR
QR = 90[255] + 36[310]
QR = (90*sin255+36*sin310) + (90*cos255+36*cos310)i
QR = -115 - 0.15i
Tan A = -115/-0.15
A = 89.9o CW.
QR/sin125 = 90/sin40
QR = 115 km.
B. PR/sin Q = PQ/sin R.
PR/sin15 = 90/sin40
PR = 36 km.
QR = QP+PR
QR = 90[255] + 36[310]
QR = (90*sin255+36*sin310) + (90*cos255+36*cos310)i
QR = -115 - 0.15i
Tan A = -115/-0.15
A = 89.9o CW.