## Trigonometry (11th Edition) Clone

$A = 9.49$
From Exercise 77: $a = \sqrt{34}$ $b = \sqrt{29}$ $c = \sqrt{13}$ We can use the law of cosines to find angle $C$ $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $cos~C = \frac{\sqrt{34}^2+\sqrt{29}^2-\sqrt{13}^2} {2~\sqrt{34}~\sqrt{29}}$ $cos~C = 0.796$ $C = arccos(0.796)$ $C = 37.2^{\circ}$ We can find the area of the triangle: $A = \frac{1}{2}ab~sin~C$ $A = \frac{1}{2}\sqrt{34}~\sqrt{29}~sin(37.2^{\circ})$ $A = 9.49$