Precalculus (6th Edition) Blitzer

$a=4.35; c=10.9; B=66.5^{\circ}$
Here, we have $A= 23.5^{\circ}$ ; $b= 10$ Since, $\tan \theta= \dfrac{Opposite}{Adjacent}$ $a= b \tan A =10 \tan 23.5^{\circ} =4.35$ Now, $c=\dfrac{b}{\cos A}=\dfrac{10}{\cos 23.5^{\circ}}=10.9$ So, $B=90^{\circ} -23.5^{\circ}=66.5^{\circ}$ Thus, $a=4.35; c=10.9; B=66.5^{\circ}$