Question

Cho tam giác ABC , tia phân giác góc B và C cắt tại I . Tính góc BIC 

A/ góc B =80 độ góc C =40.độ

B/ góc A=80 độp

Answer 1

a) \(\widehat{IBC}=\frac{1}{2}\widehat{B}=\frac{1}{2}.80^0=40^0\)

    \(\widehat{ICB}=\frac{1}{2}.\widehat{C}=\frac{1}{2}.40^0=20^0\)

\(\Rightarrow\widehat{BIC}=180^0-\widehat{IBC}-\widehat{ICB}=180^0-40^0-20^0=120^0\)

b) Ta có : \(\widehat{A}=80^0\)

\(\Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-80^0=100^0\)

\(\Rightarrow\widehat{IBC}+\widehat{ICB}=\frac{\widehat{B}+\widehat{C}}{2}=\frac{100^0}{2}=50^0\)

\(\Rightarrow\widehat{BIC}=180^0-\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-50^0=130^0\)