Cho tam giác ABC có góc B = 60 độ, góc C = 45 độ, AC = 10. Tính a, R, S, r.

Xét ΔABC, có:

Ta có: $\widehat{A}={180}^0-\widehat{B}-\widehat{C}={180}^0-{60}^0-{45}^0={75}^0.$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ (định lí sin)

$\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{10}{\sin {60}^0}=\frac{10}{\frac{\sqrt{3}}{2}}=\frac{20}{\sqrt{3}}\Rightarrow a=\frac{20}{\sqrt{3}}\sin A=\frac{20}{\sqrt{3}}.\sin {75}^0=11,15$

$\Rightarrow 2R=\frac{b}{\sin B}=\frac{20}{\sqrt{3}}\iff R=\frac{10}{\sqrt{3}}$..

Diện tích tam giác ABC là: $S=\frac{1}{2}.b.a.\sin C=\frac{1}{2}\mathrm{.10.11},15.\sin {45}^0\approx 39,42$ (đvdt)

$\Rightarrow \frac{c}{\sin C}=\frac{b}{\sin B}=\frac{20}{\sqrt{3}}\Rightarrow c=\frac{20}{\sqrt{3}}\sin C=\frac{20}{\sqrt{3}}.\sin {45}^0=\frac{10\sqrt{6}}{3}$

Ta có: $S=pr=\frac{\left(a+b+c\right)r}{2}\Rightarrow r=\frac{2S}{a+b+c}=\frac{2.39,42}{11,15+10+\frac{10\sqrt{6}}{3}}=2,69$

Vậy a = 11,15; $R=\frac{10}{\sqrt{3}},c=\frac{10\sqrt{6}}{3},$ r = 2,69.