Ta có :
\(\widehat{BDC}+\widehat{BDA}=180^0\) ( 2 góc kề bù )
Mà \(\widehat{BDC}=100^0\Rightarrow\widehat{BDA}=80^0\)
Do BD là p/g \(\widehat{B}\)
\(\Rightarrow\widehat{ABD}=\widehat{CBD}\)
\(2\widehat{A}=3\widehat{C}\)
\(\Rightarrow\widehat{A}=\dfrac{3\widehat{C}}{2}=1,5\widehat{C}\)
Lại có :
Xét tam giác ABD có góc BDC là góc ngoài đỉnh D
\(\Rightarrow\widehat{BDC}=\widehat{A}+\widehat{ABD}=1,5\widehat{C}+\widehat{ABD}=100^0\left(1\right)\)
Xét tam giác BDC có góc BDA là góc ngoài đỉnh D
\(\Rightarrow\widehat{BDA}=\widehat{CBD}+\widehat{C}=80^0\left(2\right)\)
Lấy ( 1 ) - ( 2 ) ; ta được :
\(0,5\widehat{C}=100^0-80^0\)
\(\Rightarrow0,5\widehat{C}=20^0\)
\(\Rightarrow\widehat{C}=40^0\)
\(\Rightarrow\widehat{A}=40^0.1,5=60^0\)
Mà \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) ( tổng 3 góc trong 1 t/g )
Nên : \(\widehat{B}=180^0-40^0-60^0=80^0\)
Vậy \(\widehat{A}=40^0;\widehat{B}=60^0;\widehat{C}=80^0\)
