Chapter 11 - Section 11.2 - The Cosine Ratio and Applications - Exercises - Page 504: 42

$A = 40.41~cm^2$

Since the vertex angle is $46^{\circ}$, we can let $\theta = \frac{46^{\circ}}{2} = 23^{\circ}$ We can find the area of the triangle: $A = s^2~sin~\theta~cos~\theta$ $A = (10.6~cm)^2~sin~23^{\circ}~cos~23^{\circ}$ $A = 40.41~cm^2$