Lời giải:
Tổng 3 góc trong tam giác \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
Tam giác $ABC$ cân tại $A$ nên \(\widehat{B}=\widehat{C}\)
\(\Rightarrow \widehat{B}+\widehat{C}=2\widehat{B}\)
\(\Rightarrow \) \(\widehat{A}+2\widehat{B}=180^0\Rightarrow \widehat{B}=\frac{180^0-\widehat{A}}{2}=\widehat{C}\)
Vậy, tam giác $ABC$ cân tại $A$ thì \(\widehat{B}=\widehat{C}=\frac{180^0-\widehat{A}}{2}\)
Áp dụng vào bài toán:
a)
Mà \(\widehat{B}=\widehat{C}=\frac{180^0-\widehat{A}}{2}=\frac{180^0-70^0}{2}=55^0\)
b)
\(\widehat{C}=\widehat{B}=80^0\)
\(\widehat{A}=180^0-\widehat{B}-\widehat{C}=180^0-80^0-80^0=20^0\)
c)
\(30^0=\widehat{A}-\widehat{B}=\widehat{A}-\frac{180^0-\widehat{A}}{2}=\frac{3\widehat{A}-180^0}{2}\)
\(\Rightarrow 3\widehat{A}=2.30^0+180^0=240^0\Rightarrow \widehat{A}=80^0\)
\(\Rightarrow \widehat{C}=\widehat{B}=\widehat{A}-30^0=80^0-30^0=50^0\)
d) \(\widehat{A}=2\widehat{B}=2.\frac{180^0-\widehat{A}}{2}=180^0-\widehat{A}\)
\(\Rightarrow 2\widehat{A}=180^0\Rightarrow \widehat{A}=90^0\)
\(\widehat{C}=\widehat{B}=\frac{\widehat{A}}{2}=\frac{90^0}{2}=45^0\)
e)
\(120^0=\widehat{A}+\widehat{B}=\widehat{A}+\frac{180^0-\widehat{A}}{2}=\frac{180^0+\widehat{A}}{2}\)
\(\Rightarrow \widehat{A}=60^0\)
\(\Rightarrow \widehat{C}=\widehat{B}=120^0-\widehat{A}=120^0-60^0=60^0\)
