# A triangle has sides A, B, and C. Sides A and B have lengths of 4 and 5, respectively. The angle between A and C is (7pi)/24 and the angle between B and C is  (3pi)/8. What is the area of the triangle?

Dec 15, 2015

$5 \sqrt{3}$

#### Explanation:

We use ABC for points; and a,b,c for opposite sides.

angle between a and c = $\hat{B} = \frac{7}{24} \pi$

angle between b and c = $\hat{A} = \frac{3}{8} \pi$

$\hat{C} = \pi - \hat{B} - \hat{A} = \pi \left(1 - \frac{7}{24} - \frac{3}{8}\right) = \frac{1}{3} \pi$

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \hat{C}$

${c}^{2} = 16 + 25 - 20$

$c = \sqrt{21}$

${S}_{\Delta} = \sqrt{p \left(p - a\right) \left(p - b\right) \left(p - c\right)}$

$p = \frac{9 + \sqrt{21}}{2}$

$p - 4 = \frac{1 + \sqrt{21}}{2}$

$p - 5 = \frac{- 1 + \sqrt{21}}{2}$

$p - \sqrt{21} = \frac{9 - \sqrt{21}}{2}$

${S}_{\Delta}^{2} = \frac{9 + \sqrt{21}}{2} \cdot \frac{1 + \sqrt{21}}{2} \cdot \frac{- 1 + \sqrt{21}}{2} \cdot \frac{9 - \sqrt{21}}{2}$

$16 \cdot {S}_{\Delta}^{2} = \left(81 - 21\right) \cdot \left(21 - 1\right)$

${S}_{\Delta} = \sqrt{\frac{60 \cdot 20}{16}} = \frac{10}{4} \sqrt{12} = \frac{5}{2} \cdot 2 \sqrt{3}$