Question

In triangle ABC, measure of angle A = 42, measure of angle C = 56, and a = 12. Find c to the nearest tenth.

Answer 1

`\textit{Law of Sines} \\\\ \cfrac{a}{\sin(\measuredangle A)}=\cfrac{b}{\sin(\measuredangle B)}=\cfrac{c}{\sin(\measuredangle C)} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{c}{\sin(56^o)}=\cfrac{12}{\sin(42^o)}\implies c\sin(42^o)=12\sin(56^o) \\\\\\ c=\cfrac{12\sin(56^o)}{\sin(42^o)}\implies c\approx 14.9`

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