Question

Solve and explain question

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{p}{\sin(40^o)}=\cfrac{10}{\sin(62^o)}\implies p\sin(62^o)=10\sin(40^o) \\\\\\ p=\cfrac{10\sin(40^o)}{\sin(62^o)}\implies p\approx 7.3`

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