# A triangle has sides A, B, and C. The angle between sides A and B is pi/6 and the angle between sides B and C is pi/12. If side B has a length of 5, what is the area of the triangle?

Sep 7, 2016

Area $\approx 2.2877$

#### Explanation:

If $\angle B C = \frac{\pi}{12}$ and $\angle A B = \frac{\pi}{6}$
then
$\textcolor{w h i t e}{\text{XXX}} \angle A C = \pi - \left(\frac{\pi}{12} + \frac{\pi}{6}\right) = \frac{3 \pi}{4}$

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By the Law of Sines:
$\textcolor{w h i t e}{\text{XXX}} \frac{A}{\sin} \left(\angle B C\right) = \frac{B}{\sin} \left(\angle A C\right) = \frac{C}{\sin} \left(\angle A B\right)$

With the given values:
$\textcolor{w h i t e}{\text{XXX}} \frac{A}{\sin} \left(\frac{\pi}{12}\right) = \frac{5}{\sin} \left(\frac{3 \pi}{4}\right) = \frac{C}{\sin} \left(\frac{\pi}{6}\right)$

So
$\textcolor{w h i t e}{\text{XXX}} A = \frac{5}{\sin} \left(\frac{3 \pi}{4}\right) \cdot \sin \left(\frac{\pi}{12}\right) \approx 1.830127019$
and
$\textcolor{w h i t e}{\text{XXX}} C = \frac{5}{\sin} \left(\frac{3 \pi}{4}\right) \cdot \sin \left(\frac{\pi}{6}\right) \approx 3.535533906$

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The semi-perimeter of the triangle, $s$, is
$\textcolor{w h i t e}{\text{XXX}} s = \frac{A + B + C}{2} \approx 5.182830462$

By Heron's Formula, the Area of the Triangle is
$\textcolor{w h i t e}{\text{XXX}} A = \sqrt{s \left(s - A\right) \left(s - B\right) \left(s - C\right)}$

$\textcolor{w h i t e}{\text{XXXX}} \approx 2.287658774$