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
2 answers
answered above
A)distance qr=
|ar|^2=q^2+r^2-2pqcosP
=90^2+50^2-2*50*90cos125
=8100+2500-9000cos125
=10600+5162.1879
=root of 10600+5162.1879
qr=130
b)The bearing of r from q=
=q/sinQ=p/sinP
=90/sinQ=130/sin125
=130sinQ=90sin125
Q(sin)=73.72/130
Q=34.6
Then the bearing of r from q=360-34.6
=325.4
|ar|^2=q^2+r^2-2pqcosP
=90^2+50^2-2*50*90cos125
=8100+2500-9000cos125
=10600+5162.1879
=root of 10600+5162.1879
qr=130
b)The bearing of r from q=
=q/sinQ=p/sinP
=90/sinQ=130/sin125
=130sinQ=90sin125
Q(sin)=73.72/130
Q=34.6
Then the bearing of r from q=360-34.6
=325.4
1 answer