Ta có: \(B_1+BIC+C_1=180^0\)( tổng 3 góc tam giác )
\(B_1=\frac{B}{2}=\frac{100^0}{2}=50^0\)
Hay:\(50^0+110^0+C_1=180^0\)
Vậy: \(C_1=180^0-\left(50^0+110^0\right)=20^0\)
\(2C_1=ACB=20^0.2=40^0\)
Xét \(\Delta ABC,\)ta có:
\(CAB+B+ACB=180^0\)( tổng 3 góc tam giác )
Hay:\(CAB+100^0+40^0=180^0\)
Vậy:\(CAB=180^0-\left(100^0-40^0\right)=40^0\)
\(\Rightarrow ACB=40^0,CAB=40^0\)