How to find the remaining segment of this triangle without the Law of Cosines?

Question

I am given a triangle $T$ with vertices $A, B,$ and $C$ and the the following info about $T:$

1. $\overline{AC} = 6$ Km
2. $\overline{BC} = 9$ Km
3. The angle formed by these two segments is $120^{\circ}$

I have to find $\overline{AB}.$ Now, I know one straight way to do this is using the Law of Cosines. My question is, is there any other way to solve this problem without the Law of Cosines?

Answer 1

Extend $\overline{AC}$ past $C$ and drop perpendicular $\overline{BD}$ where $D$ lies on the extension.

You can determine the length of $\overline{CD}$ and $\overline{BD}$ as $\left||\overline{BC}|\cos 120^{\circ}\right|$ and $\left||\overline{BC}|\sin 120^{\circ}\right|$, respectively.

Then, $|\overline{AB}|$ as $\sqrt{(|\overline{AC}| + |\overline{CD}|)^2 + |\overline{BD}|^2}$.

Answer 2

Consider the diagram below.

Draw altitude $\overline{AD}$ to the extension of side $\overline{BC}$ to form right triangle $ABD$.

Since $\angle ACD$ is supplementary to $\angle ACB$, its measure is $180^\circ - 120^\circ = 60^\circ$.

Thus, $$|\overline{CD}| = 6~\text{km}\cos(60^\circ) = 3~\text{km}$$ Hence, $$|\overline{BD}| = |\overline{BC}| + |\overline{CD}| = 6~\text{km} + 3~\text{km} = 9~\text{km}$$ and $$|\overline{AD}| = 6~\text{km}\sin(60^\circ) = 3\sqrt{3}~\text{km}$$ To find $|\overline{AB}|$, use the Pythagorean Theorem on right triangle $ABD$.