Question

solve for the remaining angle and sides of the triangle described below. round to the nearest hundredth: (a = 40^{circ},b = 60^{circ},a = 3)

Explanation:

Step1: Find angle C

The sum of angles in a triangle is $180^{\circ}$. So $C=180^{\circ}-A - B$.
$C = 180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}$

Step2: Use the Law of Sines to find side b

The Law of Sines states $\frac{a}{\sin A}=\frac{b}{\sin B}$.
So $b=\frac{a\sin B}{\sin A}$.
$b=\frac{3\times\sin60^{\circ}}{\sin40^{\circ}}=\frac{3\times\frac{\sqrt{3}}{2}}{\sin40^{\circ}}\approx\frac{3\times0.8660}{0.6428}\approx4.02$

Step3: Use the Law of Sines to find side c

Using the Law of Sines $\frac{a}{\sin A}=\frac{c}{\sin C}$.
So $c=\frac{a\sin C}{\sin A}$.
$c=\frac{3\times\sin80^{\circ}}{\sin40^{\circ}}=\frac{3\times0.9848}{0.6428}\approx4.59$

Answer:

$C = 80^{\circ}, b\approx4.02, c\approx4.59$