Answer
$b\approx295.95$
$\angle A\approx6.64^{\circ}$
Work Step by Step
From law of cosines, we have
$b^{2}=a^{2}+c^{2}-2ac\cos B$
$=(100)^{2}+(200)^{2}-2(100)(200)\cos 160^{\circ}$
$\approx87587.7048$
$b\approx\sqrt {87587.7048}$
$b\approx295.95$
From law of sines, we have
$\frac{\sin A}{a}=\frac{\sin B}{b}$
$\implies \sin A=\frac{a}{b} \sin B=\frac{100}{295.95}\times\sin 160^{\circ}$
$\approx0.115566867$
$\angle A\approx\arcsin (0.115566867)$
$\angle A\approx6.64^{\circ}$
