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Question #93300
Two ships P and Q left a port R at the same time on different routes. Q sailed on a bearing of 150° while P sailed on the north side of Q.After a distance of 8km and 10 km by p anq respectively their distance apart was 12km. Find the bearing of P from R.
Expert's answer
Find the angle
x
x
x
between RP and RQ using the law of cosines:
P
Q
2
=
R
P
2
+
R
Q
2
−
2
R
P
⋅
R
Q
⋅
c
o
s
x
PQ^2=RP^2+RQ^2-2RP\cdot RQ\cdot cos x
P
Q
2
=
R
P
2
+
R
Q
2
−
2
RP
⋅
RQ
⋅
cos
x
c
o
s
x
=
R
P
2
+
R
Q
2
−
P
Q
2
2
R
P
⋅
R
Q
cos x=\frac{RP^2+RQ^2-PQ^2}{2RP\cdot RQ}
cos
x
=
2
RP
⋅
RQ
R
P
2
+
R
Q
2
−
P
Q
2
c
o
s
x
=
8
2
+
1
0
2
−
1
2
2
2
⋅
8
⋅
10
=
64
+
100
−
144
160
=
20
160
=
1
8
cos x=\frac{8^2+10^2-12^2}{2\cdot 8\cdot 10}=\frac{64+100-144}{160}=\frac{20}{160}=\frac{1}{8}
cos
x
=
2
⋅
8
⋅
10
8
2
+
1
0
2
−
1
2
2
=
160
64
+
100
−
144
=
160
20
=
8
1
x
≈
83
°
x\approx 83\degree
x
≈
83°
Then the bearing of P from R is:
150
°
−
83
°
=
67
°
150\degree-83\degree=67\degree
150°
−
83°
=
67°
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