## Algebra 2 (1st Edition)

$A=37.6^{\circ}; B=38.4^{\circ}, a= 15.7$
Law of sines: $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$ Here, $\dfrac{\sin 104^{\circ}}{25}=\dfrac{\sin B}{16}$ This implies: $B=38.4^{\circ}$ Since a triangle always has $180^{\circ}$, we need to add the known angles together and subtract the result from $180^{\circ}$. Thus, $A=180^{\circ}-(104^{\circ}+38.4^{\circ})=37.6^{\circ}$ Now, $\dfrac{\sin 104^{\circ}}{25}=\dfrac{\sin 37.6^{\circ}}{a}$ $\implies a=\dfrac{25 \sin 37.6^{\circ}}{\sin 104^ {\circ}}=15.7$ Hence, $A=37.6^{\circ}; B=38.4^{\circ}, a= 15.7$