Hello, yuka!
A ship sails on a course of 038° from an island.
After 50km it changes course to 140° and sails for another 100km.
Find: a) the distance from the boat to the island
\(\displaystyle \;\;\;\,\) b) the bearing from the island to the boat
Code:
P Q
: :
: B:
: * 140°
: /:\
: / : \
: / : \
:38°/38°:40°\
: / : \
: /50 : \100
:/ : \
A* : \
: * : \
: * \
: : * \
: : * \
: R *C
We are given: \(\displaystyle \angle PAB\,=\,38^o\) . . . Hence: \(\displaystyle \,\angle ABR\,=\,38^o\)
\(\displaystyle \;\;\)and \(\displaystyle AB\,=\,50\) km.
We are given: \(\displaystyle \angle QBC\,=\,140^o\) . . . Hence: \(\displaystyle \angle RBC\,=\,40^o\)
\(\displaystyle \;\;\)and \(\displaystyle BC\,=\,100\) km.
In \(\displaystyle \Delta ABC\), we have: \(\displaystyle AB\,=\,50,\:BC\,=\,100,\;\angle B\,=\,78^o\)
Law of Cosines: \(\displaystyle \,AC^2\:=\:50^2\,+\,100^2\,-\,2(50)(100)\cos78^o\:=\:10420.88309\)
a) Therefore: \(\displaystyle \,AC\:=\:102.0827267\:\approx\:102\) km
The ship's bearing is: \(\displaystyle \angle PAC\:=\:38^o\,+\,\angle BAC\:=\:38^o\,+\,\angle A\)
In \(\displaystyle \Delta ABC\), we have: \(\displaystyle AB\,=\,50,\:AC\,=\,102,\:BC\,=\,100\)
Law of Cosines: \(\displaystyle \:\cos A\:=\:\frac{50^2\,+\,102^2\,-\,100^2}{2(50)(102)}\:=\:0.284705882\)
Hence: \(\displaystyle \,\angle A\:=\:73.45873143\:\approx\;73.5^o\)
b) Therefore, the ship's bearing is: \(\displaystyle \,38^o\,+\,73.5^o\:=\:111.5^o\)
