Monday

July 28, 2014

July 28, 2014

Posted by **bearing question corrected** on Thursday, January 7, 2010 at 1:29pm.

(a)distance between q and r

(b)bearing of r from q

- bearing -
**bobpursley**, Thursday, January 7, 2010 at 3:59pmanswered above

- bearing -
**Sani mohammed**, Sunday, April 14, 2013 at 4:50amA)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

- bearing -
**Sani mohammed**, Sunday, April 14, 2013 at 4:50amA)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

- bearing -
**Sani mohammed**, Sunday, April 14, 2013 at 4:51amA)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

- bearing -
**Sani**, Sunday, April 14, 2013 at 4:51am

|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

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