Ok, the angle at p is between 310 and 075,or an angle of 125 (check that). YOu have two sides, one angle
find another angle with the law of sines.
then find the last angle (sum of angles is 180)
then find the last side by law of sines.
Make certain you draw the figure, life is simpler with a drawing.
Three towns p, q and r are such that the distance between p and q is 50km and the distance between p and r is 90km. If the bearing of q from p 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
4 answers
thank you, although am still hooked
254
All angles are measured CW from +y-axis
a. PQ + QR = PR.
QR = PR - PQ = 90[310o] - 50[75o],
QR = (90*sin310-50*sin75) + (90*cos310-50*cos75)i,
QR = -117.2 + 44.9i = 125.5km[-69o] = 125.5km[291o] CW.
b. Bearing = 291o CW.
a. PQ + QR = PR.
QR = PR - PQ = 90[310o] - 50[75o],
QR = (90*sin310-50*sin75) + (90*cos310-50*cos75)i,
QR = -117.2 + 44.9i = 125.5km[-69o] = 125.5km[291o] CW.
b. Bearing = 291o CW.
1 answer