Trigonometry

[yuka]

how do I do this? :roll:

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
b) the bearing from the island to the boat

 

[soroban]

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

      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: $\displaystyle \,38^o\,+\,73.5^o\:=\:111.5^o$