Chapter 11 - Section 11.2 - The Cosine Ratio and Applications - Exercises - Page 504: 43

$A = 6.77~in^2$

Since the measure of a base angle is $72^{\circ}$, we can let $\theta = 180^{\circ}-72^{\circ}-90^{\circ} = 18^{\circ}$ We can find the area of the triangle: $A = s^2~sin~\theta~cos~\theta$ $A = (4.8~in)^2~sin~18^{\circ}~cos~18^{\circ}$ $A = 6.77~in^2$