## Algebra 2 (1st Edition)

$A=72^{\circ}\\ C=90^{\circ}\\ a \approx 22.8\\b \approx 7.4$
$\sin 18^{\circ}=\dfrac{b}{c}=\dfrac{b}{24}$ This gives: $b \approx 7.4$ and $\cos 18^{\circ}=\dfrac{a}{c}=\dfrac{a}{24}$ This gives: $a \approx 22.8$ Since the sum of a triangle is $180^{\circ}$ and the right angle is $90^{\circ}$ $A=B-C=180^{\circ}-90^{\circ}-18^{\circ}=72^{\circ}$