# A triangle has sides A, B, and C. The angle between sides A and B is (7pi)/12. If side C has a length of 1  and the angle between sides B and C is pi/12, what is the length of side A?

Jan 25, 2016

The length of side $A$ is $2 - \sqrt{3}$.

#### Explanation:

You can use the law of sines.

The angle between sides $A$ and $B$ is the angle opposing the side $C$. Let's call this angle $\gamma$.

Similarly, the angle between sides $B$ and $C$ is the one opposing the side $A$. Let's call this angle $\alpha$.

Even though we don't need it for this particular task, let's also call the last remaining angle $\beta$, this one is opposing the side $B$ .

According to the law of sines, the following relation between the sides and the opposite angles exists:

$\frac{A}{\sin} \alpha = \frac{B}{\sin} \beta = \frac{C}{\sin} \gamma$

In our case, we only need to look at $A$, $C$, $\alpha$ and $\gamma$:

$\frac{A}{\sin} \alpha = \frac{C}{\sin} \gamma$

$\frac{A}{\sin} \left(\frac{\pi}{12}\right) = \frac{1}{\sin} \left(\frac{7 \pi}{12}\right)$

$\iff A = \sin \frac{\frac{\pi}{12}}{\sin} \left(\frac{7 \pi}{12}\right)$

So, the only thing left to do is compute $\sin \left(\frac{\pi}{12}\right)$ and $\sin \left(\frac{7 \pi}{12}\right)$.

Let me show you how to do this without the calculator but with some basic knowledge about $\sin$ and $\cos$ and using the identity

$\sin \left(a - b\right) = \sin \left(a\right) \cos \left(b\right) - \cos \left(a\right) \sin \left(b\right)$.

You need to express $\frac{\pi}{12}$ as a sum or difference of simpler values:

$\sin \left(\frac{\pi}{12}\right) = \sin \left(\frac{\pi}{3} - \frac{\pi}{4}\right)$

$= \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) - \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$

$= \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{1}{\sqrt{2}}$

$= \frac{1}{2 \sqrt{2}} \left(\sqrt{3} - 1\right)$

Similarly,

$\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{3} + \frac{\pi}{4}\right)$

$= \sin \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{4}\right)$

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$

$= \frac{1}{2 \sqrt{2}} \left(\sqrt{3} + 1\right)$

Thus, the length of the side $A$ is

$A = \sin \frac{\frac{\pi}{12}}{\sin} \left(\frac{7 \pi}{12}\right)$

$= \frac{\frac{1}{2 \sqrt{2}} \left(\sqrt{3} - 1\right)}{\frac{1}{2 \sqrt{2}} \left(\sqrt{3} + 1\right)}$

$= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

$= \frac{\left(\sqrt{3} - 1\right) \textcolor{b l u e}{\left(\sqrt{3} - 1\right)}}{\left(\sqrt{3} + 1\right) \textcolor{b l u e}{\left(\sqrt{3} - 1\right)}}$

$= {\left(\sqrt{3} - 1\right)}^{2} / \left({\left(\sqrt{3}\right)}^{2} - {1}^{2}\right)$

$= \frac{3 - 2 \sqrt{3} + 1}{2}$

$= 2 - \sqrt{3}$