A triangle has sides A, B, and C. Sides A and B have lengths of 2 and 6, respectively. The angle between A and C is (pi)/8 and the angle between B and C is  (3pi)/4. What is the area of the triangle?

Jan 27, 2018

Area $= 6 \sin \left(\frac{\pi}{8}\right) = 3 \sqrt{2 - \sqrt{2}} \approx 2.296$.

Explanation:

We know $a = 2$, $b = 6$, $B = \frac{\pi}{8}$, and $A = \frac{3 \pi}{4}$.

We can calculate $C = \pi - \left(\frac{\pi}{8} + \frac{3 \pi}{4}\right) = \frac{\pi}{8}$.

The area of the triangle is $\frac{1}{2} a \cdot b \cdot \sin \left(C\right)$, so:

Area $= \frac{1}{2} \left(2\right) \left(6\right) \sin \left(\frac{\pi}{8}\right) = 6 \sin \left(\frac{\pi}{8}\right)$

You can find $\sin \left(\frac{\pi}{8}\right)$ using the half-angle formula for sine:

$\sin \left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos \left(x\right)}{2}}$

So $\sin \left(\frac{\pi}{8}\right) = \sqrt{\frac{1 - \cos \left(\frac{\pi}{4}\right)}{2}}$

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

so, back to the area:

Area $= 6 \sin \left(\frac{\pi}{8}\right) = 6 \frac{\sqrt{2 - \sqrt{2}}}{2} = 3 \sqrt{2 - \sqrt{2}}$

or Area $\approx 2.296$