Given:
Angle A = 120 degrees
Side opposite angle C = 150 meters
Side opposite angle B = 275 meters
Find:
Angle B
Solution:
Since we have two sides given and an included angle, we can use cosine law.
Let's look for the length of the side opposite Angle A first.
[tex]a^2=b^2+c^2-2bc\cos A[/tex]where a = length of the side opposite Angle A or side BC
b = side opposite Angle B or Side AC
c = side opposite Angle C or Side AB
A = Angle A
Since we already have the data above, let's plug it in to the formula we have.
[tex]a^2=275^2+150^2-2(275)(150)\cos 120[/tex]Then, solve a.
[tex]\begin{gathered} a^2=75,625+22,500-82,500(-0.5) \\ a^2=98,125+41,250 \\ a^2=139,375 \\ \sqrt[]{a^2}=\sqrt[]{139,375} \\ a\approx373.3296\approx373.33 \end{gathered}[/tex]Hence, the length of side opposite a or Side BC is approximately 373.33 meters.
Now, to solve for Angle B, we can use the sine law.
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}[/tex]Let's plug in the value of Angle A, side BC or a, and side AC or b to the formula.
[tex]\frac{\sin120}{373.3296}=\frac{\sin B}{275}[/tex]Then, solve for Angle B.
[tex]\begin{gathered} \sin B=\frac{275\sin 120}{373.3296} \\ \sin B=0.6379268552 \\ B=\sin ^{-1}0.6379268552 \\ B\approx39.64 \end{gathered}[/tex]Therefore, the bearing of ship C from ship B is approximately 40 degrees. (rounded off to the nearest degree)
