Question

Using the Law of Sines, if m∠A is 66°, m∠B is 53°, and side c = 23km, what is side b?

Answer 1

`\bf \begin{cases} \measuredangle A=66^o\\ \measuredangle B=53^o\\ \textit{what about the last angle? C?}\\ \textit{all internal angles in a triangle add up to 180, thus}\\ \measuredangle C = 180-(A+B)\to 61^o \\\\ \textit{side`
`\bf thus \\\\ \cfrac{sin(\measuredangle C)}{c}=\cfrac{sin(B)}{b}\implies \cfrac{sin(61^o)}{23}=\cfrac{sin(53^o)}{b}`

solve for "b",

when taking the sines, make sure your calculator is in Degree mode, since the angles are in degrees