# A triangle has sides A, B, and C. Sides A and B have lengths of 8 and 15, respectively. The angle between A and C is (7pi)/24 and the angle between B and C is  (5pi)/8. What is the area of the triangle?

Nov 29, 2016

If $\alpha , \beta$ and $\gamma$ are tha angles opposite to $A , B$ and $C$, the theorem of sines states that:

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

We know that:

$\alpha = \frac{5 \pi}{8}$

$\beta = \frac{7 \pi}{24}$

and as the sum of the internal angles of a triangle always equals $\pi$

$\gamma = \pi - \alpha - \beta = \frac{\pi}{12}$

We can use this to determine $C$:

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

Then we can use Eron's formula to calculate the area from the sides:

$S = \sqrt{p \left(p - A\right) \left(p - B\right) \left(p - C\right)}$

where $p = \frac{A + B + C}{2}$

As the angles are not such as the sine is immediately known we can use approximate values or do a bit of computation using trigonometric formulas.

$\sin \alpha = \sin \left(\frac{5 \pi}{8}\right) = \sin \left(\frac{1}{2} \cdot \frac{5 \pi}{4}\right) = \sqrt{\frac{1 - \cos \left(\frac{5 \pi}{4}\right)}{2}} = \sqrt{\frac{1 + \cos \left(\frac{\pi}{4}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}$

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

Then:

$A = 8$
$B = 15$
$C = 8 \cdot \frac{\cancel{2}}{\sqrt{2 + \sqrt{2}}} \cdot \left(\frac{\sqrt{3} - 1}{\cancel{2} \sqrt{2}}\right)$