Theo định lí sin đối với tam giác ABC ta có:
\(\dfrac{{BC}}{{{\mathop{\rm sinA}\nolimits} }} = \dfrac{{AB}}{{{\mathop{\rm sinC}\nolimits} }} \Leftrightarrow \dfrac{5}{{\sin A}} = \dfrac{{12}}{{\sin {{37}^0}}}\)\( \Rightarrow \sin A = \dfrac{{5.\sin {{37}^0}}}{{12}} \approx 0,2508\)
\(\widehat A \approx {14^0}31'\)
\(\widehat B \approx {180^0} - \left( {{{37}^0} + {{14}^0}31'} \right) = {128^0}29'\)
\(\dfrac{{AC}}{{\sin B}} = \dfrac{{12}}{{{\mathop{\rm sinC}\nolimits} }}\)\( \Rightarrow AC = \dfrac{{12\sin B}}{{\sin C}}\) \( \approx \dfrac{{12.\sin {{128}^0}29'}}{{\sin {{37}^0}}} \approx 15,61(m)\)
Vậy khoảng cách \(AC \approx 15,61(m)\).
