Cho tam giác ABC có BC = 10 và 5 sin A = 4 sin B = 6 sin C. Khi đó:

DSDD

Ta có \(LCM(4;5;6) = 60\). Suy ra:

\(5\sin A = 4\sin B = 6\sin C \Leftrightarrow \frac{{\sin A}}{{12}} = \frac{{\sin B}}{{15}} = \frac{{\sin C}}{{10}} = k\).

Do \(BC = 10\), theo định lý sin: \(\frac{{AC}}{{\sin B}} = \frac{{AB}}{{\sin C}} = \frac{{BC}}{{\sin A}} = 2R\).

\(AC = \frac{{10\sin B}}{{\sin A}} = 10.\frac{{15k}}{{12k}} = 12.5\)

\(AB = \frac{{10\sin C}}{{\sin A}} = 10 \cdot \frac{{10k}}{{12k}} = \frac{{25}}{3}\)

Chu vi tam giác \(ABC = AB + BC + AC = 10 + 12.5 + \frac{{25}}{3} = \frac{{185}}{6}\).

\(\cos A = \frac{{A{B^2} + A{C^2} - B{C^2}}}{{2AB \cdot AC}} = \frac{{{{(\frac{{25}}{3})}^2} + {{12.5}^2} - {{10}^2}}}{{2 \cdot \frac{{25}}{3} \cdot 12.5}} = \frac{{181}}{{300}}\)

\(p = \frac{{AB + BC + AC}}{2} = \frac{{185}}{{12}}\)

\(S = \sqrt {p(p - AB)(p - BC)(p - AC)}  = \sqrt {\frac{{185}}{{12}}(\frac{{185}}{{12}} - \frac{{25}}{3})(\frac{{185}}{{12}} - 10)(\frac{{185}}{{12}} - 12.5)}  \approx 41.5\)