Question

How would I solve this?

Answer 1

`\textit{Law of Sines} \\\\ \cfrac{\sin(\measuredangle A)}{a}=\cfrac{\sin(\measuredangle B)}{b}=\cfrac{\sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\sin( x )}{17}=\cfrac{\sin( 43^o )}{16}\implies 16\sin(x)=17\sin(43^o) \implies \sin(x)=\cfrac{17\sin(43^o)}{16} \\\\\\ x=\sin^(-1)\left( ~~ \cfrac{17\sin( 43^o)}{16} ~~\right)\implies x\approx 46.44^o`

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