Precalculus (6th Edition) Blitzer

$a=3.79$ ; $b=9.25$ and $\angle B=67.7^{\circ}$
The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ $\sin 22.3^{\circ}=\dfrac{a}{10} \implies a=3.79$ and $b= 10 \cos A=10 \cos 22.3^{\circ}=9.25$ So, $\angle B=90^{\circ}- 22.3^{\circ}=67.7^{\circ}$ Thus, $a=3.79$ ; $b=9.25$ and $\angle B=67.7^{\circ}$