$\widehat{A}=180^{o} - \widehat{B}-\widehat{C}=75^{o}$
Áp dụng định lí sin trong tam giác ABC có: $\frac{BC}{sinA}=\frac{AC}{sin B}=\frac{AB}{sin C}$
hay: $\frac{BC}{sin75^{o}}=\frac{10}{sin 60^{o}}=\frac{AB}{sin 45^{o}}$
$\Rightarrow BC = a\approx 11,1$; $AB = c\approx 8,2 $
$R = \frac{a}{2sin A}\approx 5,8$
$S = \frac{1}{2}BC.AC.sinC\approx 39$
$r=\frac{2S}{AB+BC+AC}\approx 2,7$