Trong △ ABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^O\) (định lý)
=>\(30^o+\widehat{B}+30^O=180^o\)
=> \(\widehat{B}=120^o\)
Vì BD là tia phân giác góc B
=> \(\widehat{ABD}=\widehat{CBD}=\dfrac{1}{2}\widehat{B}=\dfrac{1}{2}.120^o=60^o\)
Trong △ BCD có \(\widehat{CBD}+\widehat{C}+\widehat{BDC}=180^o\) (định lý)
=> \(60^o+30^o+\widehat{BDC}=180^o\)
=> \(\widehat{BDC}=90^o\)