# A triangle has two corners with angles of  pi / 4  and  pi / 3 . If one side of the triangle has a length of 8 , what is the largest possible area of the triangle?

Jun 26, 2016

21.85252

#### Explanation:

Given for $\triangle A B C$, $\angle B = \frac{\pi}{4} , \angle C = \frac{\pi}{3}$.

We know that sum of the 3 angles of a triangle is $\pi$. Hence $\angle A = \pi - \frac{\pi}{3} - \frac{\pi}{4}$ => $\angle A = \frac{5 \pi}{12}$

As the angles $\angle B < \angle C < \angle A$, length of side b < side c < side A

To get the largest possible area of the triangle with any one side with 8, we must have the shortest side as 8.

Then b must be 8

$\frac{b}{S} \in B$ = $\frac{8}{S} \in \left(\frac{\pi}{4}\right)$ = $\frac{8}{\frac{\sqrt{2}}{2}}$ =$\frac{16}{\sqrt{2}}$ = $8 \sqrt{2}$

Using the law of Sines,

$\frac{b}{S} \in B = \frac{a}{S} \in A = \frac{c}{S} \in C$

a = $\left(\frac{b}{S} \in B\right) \cdot S \in A$ = $8 \sqrt{2}$ * $\sin \left(\frac{5 \pi}{12}\right)$ =

c= $\left(\frac{b}{S} \in B\right) \cdot S \in C$ = $8 \sqrt{2}$ * $\sin \left(\frac{\pi}{3}\right)$

Area of the triangle = $\cdot \frac{a \cdot b \cdot S \in C}{2}$

= $8 \sqrt{2}$ * $\sin \left(\frac{5 \pi}{12}\right)$ 8$\sin \left(\frac{\pi}{3}\right)$

= 21.85252