A town Q is in a bearing of 210° from P,R is another town on a bearing of 150° from P east of Q.The distance between R and P is10km .find the distance between R and Q

Answer 1

If origin of x y axis system at P
Q is 210 -180 = 30 degrees above -y axis
R is 90 + 60 degrees or 60 below +x direction from Q
that means angle PQR is ALSO 30 deg !
so QR = RP = 10

Answer 2

Given: PQ = 210o, PR = 10km[150o], QR = ?.
210-150 = 60o = Angle between given vectors.

Law of sine: RQ/sin60 = 10/sin60
QR = 10*sin60/sin60 = 10 km.

Answer 3

Interesting and satisfactory

Answer 4

P=60°,Q=60°,R=60°
PR=10km
PR= ?
Using sine rules p/sinP= q/sinQ= r/sinR
P/sin60°= 10/sin60° cross multiply
After the working
QR is equal to 10km.

Answer 5

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Answer 6

A town q is on a bearing of 210degree from p, R is another town on a bearing of 150degree from p east of q, the distance between R and p is 10km, find the distance between R and q