How to use the law of cosines Can someone please explain to me in very small, simple words and steps how to solve a triangle using the law of cosines?
 A: There are two basic situations in which you solve a triangle using the Law of Cosines.


*

*When you know all three sides $a,\,b,\,c$

*When you know two sides and the angle between them.


In the first case you must find the three angles using the angle version of the Law of Cosines
\begin{eqnarray}
\cos\alpha&=&\frac{b^2+c^2-a^2}{2bc}\\
\cos\beta&=&\frac{a^2+c^2-b^2}{2bc}\\
\cos\gamma&=&\frac{a^2+b^2-c^2}{2bc}
\end{eqnarray}
Recall, that in order for three sides to actually form a triangle, the sum of the two smaller sides must be larger than the largest sides.
The second situation in which you use the Law of Cosines is when you know two sides and the angle between them and need to find the third side.
Suppose $s_1$ and $s_2$ are the two known sides, $\theta$ is the angle between them and $s_3$ is the unknown side. Then you need the second version of the Law of Cosines to find $s_3$.
$$ s_3^2=s_1^2+s_2^2-2s_1s_2\cos\theta $$
Once you know $s_3$ you know all three sides and can find the angles using the first version of the Law of Cosines, or you may use the Law of Sines.
A: Suppose you know two sides $a$ and $b$ and the angle $\gamma$ between them. Then the law of cosines gives you the third side:
$$
c^2=a^2+b^2-2ab\cos\gamma
$$
Once you know this you can also find the angles $\alpha$ and $\beta$ (opposite to $a$ and $b$) with the law of sines:
$$
\sin\alpha=\frac{a}{c}\sin\gamma
\qquad
\sin\beta=\frac{b}{c}\sin\gamma
$$
If you know the three sides, then you can compute one angle with the law of cosines
$$
\cos\gamma=\frac{a^2+b^2-c^2}{2ab}
$$
and then the other two angles as before. The relation above is not the best for computations, but one can get a better one:
$$
1+\cos\gamma=\frac{(a+b)^2-c^2}{2ab}=\frac{(a+b+c)(a+b-c)}{2ab}
$$
so, remembering that
$$
\cos^2\frac{\gamma}{2}=\frac{1+\cos\gamma}{2}
$$
and setting $2p=a+b+c$, we get
$$
\cos\frac{\gamma}{2}=\sqrt{\frac{p(p-c)}{ab}}
$$
which in the olden times was used in order to exploit logarithms and minimize approximations errors, so improving the accuracy. Note that $\gamma/2$ is always acute, so its cosine is positive.
The similar relations obtained by permuting the letters can be used for computing (half of) the other angles.
If you instead know one side $a$ and the angles $\beta$ and $\gamma$ adjacent to it, you just use the law of sines:
$$
\alpha=\pi-(\beta+\gamma)
\qquad
b=a\frac{\sin\beta}{\sin\alpha}
\qquad
c=a\frac{\sin\gamma}{\sin\alpha}
$$
A: The law of cosines is a generalization of Pythagorean theorem used to find the length of $c$ side by:
$$c^2 = a^2 + b^2 -2\!\cdot a \cdot b \cdot \cos\theta$$
where $\theta$ is the angle between the sides $a$ and $b$.
Thus if we know three among a,b,c and $\theta$ we can find the fourth unknown.
A: If by "solving a triangle" you mean finding all sides and angles of a triangle. Then, consider triangle ABC. And suppose you know AB=c, BC=a, angle ABC=$\theta$. Then by law of cosines you can find AC. 
$$AC^2 = c^2 + b^2 - 2*b*c*cos(\theta)$$
Now you know all sides. From law of cosines we can derive:
$$cos(ABC)=(AB^2 + BC^2 - AC^2)/(2*AB*BC)$$
Then by knowing all sides you can know all angles.
So, you solved the triangle.
