# A triangle has sides A, B, and C. The angle between sides A and B is pi/6. If side C has a length of 25  and the angle between sides B and C is pi/12, what is the length of side A?

May 5, 2017

#### Explanation:

In general, the Law of Sine states that if a, b, and c are the lengths of the sides opposite angles $\alpha , \beta , \gamma$ (in that order), then
$\frac{a}{\sin} \left(\alpha\right) = \frac{b}{\sin} \left(\beta\right) = \frac{c}{\sin} \left(\gamma\right)$.

We only use the Law one pair of angles at a time. In this case, we know that the angle between A and B is $\gamma$, and the angle between B and C is $\alpha$. Therefore,

a = unknown
c = 25
$\alpha = \frac{\pi}{12}$
$\gamma = \frac{\pi}{6}$

Using the Law:

$\frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{25}{\sin} \left(\frac{\pi}{6}\right)$

$\frac{\pi}{6}$ is one of the standard angles. $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

The value of $\sin \left(\frac{\pi}{12}\right)$ may be found using the difference formula for sine:

$\sin \left(\frac{\pi}{12}\right) = \sin \left(\frac{3 \pi}{12} - \frac{2 \pi}{12}\right)$
$= \sin \left(\frac{3 \pi}{12}\right) \cos \left(\frac{2 \pi}{12}\right) - \cos \left(\frac{3 \pi}{12}\right) \sin \left(\frac{2 \pi}{12}\right)$
$= \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) - \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right)$
$= \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)$
$= \frac{\sqrt{6} - \sqrt{2}}{4}$

Solving the proportion:

$\frac{a}{\sin} \left(\frac{\pi}{12}\right) = \frac{25}{\sin} \left(\frac{\pi}{6}\right)$

$a = \frac{\left(25\right) \left(\sin \left(\frac{\pi}{12}\right)\right)}{\sin} \left(\frac{\pi}{6}\right)$

$a = \frac{\left(25\right) \frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{1}{2}}$

$a = \frac{\left(25\right) \left(\sqrt{6} - \sqrt{2}\right)}{2}$

$a = \frac{25 \left(\sqrt{6} - \sqrt{2}\right)}{2}$

NOTE: If we use the half-angle formula for sine instead of the difference formula, we obtain an answer that looks different but is equal to the above.