\(\widehat{ADB}=90^0-20^0=70^0\)
=>góc ADC=110 độ
\(x=\dfrac{180^0-110^0}{2}=35^0\)
`hatD = 180^o - hat B - hat (ABD) = 180^o - 90^o - 20^o = 70^o`
`=> hat(ADC) = 180^o - 70^o = 110^o = 2x`.
`=> x = 110^o :2 = 55^o`
Ta có:
\(\widehat{ADC}\) là góc ngoài của \(\Delta ABD\) tại D
\(\Rightarrow\widehat{ADC}=\widehat{ABD}+\widehat{BAD}=20^0+90^0=110^0\)
\(\Delta ADC\) có:
\(\widehat{ADC}+\widehat{ACD}+\widehat{CAD}=180^0\) (tổng ba góc trong \(\Delta ADC\))
\(\Rightarrow\widehat{110}^0+x+x=180^0\)
\(2x=70^0\)
\(x=35^0\)