Question

Solve the right triangle where $\angle C={90}^{\circ },b=4.5,\angle A={39}^{\circ }$. (Given that $\sin {51}^{\circ }=0.7771$ and $\sin {39}^{\circ }=0.6293$)

Answer 1

$b=4.5$

$\sin \left(51\right)=\frac{AC}{AB}=\frac{b}{c}$

$c=\frac{4.5}{\sin \left(51\right)}=\frac{4.5}{0.7771}$

$c=5.79$

$\sin \left(39\right)=\frac{BC}{AB}=\frac{BC}{C}=\frac{a}{c}$

$a=\left(0.6293\right)\left(5.79\right)$

$a=3.64$