Answer
$A = 9.49$
Work Step by Step
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:
$A = \frac{1}{2}ab~sin~C$
$A = \frac{1}{2}\sqrt{34}~\sqrt{29}~sin(37.2^{\circ})$
$A = 9.49$
