Ta có: ΔABC cân tại A (có AB = AC)
mà \(\widehat{A}=30^o\)
=> \(\widehat{B}_1=\widehat{C_1}=\dfrac{180^o-30^o}{2}=75^o\)
Ta thấy: \(\widehat{B_{12}}=90^o\)
=> \(\widehat{B_2}=90^o-75^o=15^o\)
Có: \(\widehat{C_1}=\widehat{B_2}+\widehat{N}\) (góc ngoài tam giác ABC)
=> \(\widehat{N}=\widehat{C_1}-\widehat{B_2}\)
=> \(\widehat{N}=75^o-15^o=50^o\)
Xét ΔBCN có: \(\widehat{B_2}+\widehat{C_2}+\widehat{N_1}=180^o\) (tổng 3 góc trong 1 tam giác)
=> \(\widehat{C_2}=180^o-\left(\widehat{B_2}+\widehat{N_1}\right)\)
=> \(\widehat{C_2}=180^o-\left(15^o+50^o\right)=180^o-75^o=115^o\)
Vậy \(\widehat{C_2}=115^o\)