# A triangle has sides A, B, and C. The angle between sides A and B is (pi)/3 and the angle between sides B and C is pi/6. If side B has a length of 13, what is the area of the triangle?

Dec 26, 2016

${A}_{\triangle} = \frac{169 \sqrt{3}}{8} \approx 36.59$.

#### Explanation:

Our goal will be to use ${A}_{\triangle} = \frac{1}{2} a b \sin C$. We know $b = 13$ and $\angle C = \frac{\pi}{3}$, so we need to find $a$.

Step 1: Find the value of $\angle B$.

Using the fact that the sum of all 3 angles in a triangle is $\pi$, we get

$\angle A + \angle B + \angle C = \pi$
$\frac{\pi}{6} \text{ "+ angle B + pi / 3" } = \pi$
$\text{ "angle B " } = \frac{\pi}{2}$

So $\angle B = \frac{\pi}{2}$.

Step 2: Find the length of $a$.

We now use the sine law for triangles to get

$\frac{a}{\sin} A = \frac{b}{\sin} B$

$\frac{a}{\sin} \left(\frac{\pi}{6}\right) = \frac{13}{\sin} \left(\frac{\pi}{2}\right)$

$\text{ "a" } = \frac{13 \sin \left(\frac{\pi}{6}\right)}{\sin} \left(\frac{\pi}{2}\right)$

$\text{ "a" } = \frac{13 \left(\frac{1}{2}\right)}{1} = \frac{13}{2}$

So $a = \frac{13}{2}$.

Step 3: Find the area of the triangle.

We can now use the following formula for a triangle's area:

${A}_{\triangle} = \frac{1}{2} a b \sin C$

${A}_{\triangle} = \frac{1}{2} \cdot \frac{13}{2} \cdot 13 \cdot \sin \left(\frac{\pi}{3}\right)$

${A}_{\triangle} = \frac{169}{4} \cdot \frac{\sqrt{3}}{2}$

${A}_{\triangle} = \frac{169 \sqrt{3}}{8} \text{ } \approx 36.59$.