a) Ta có: \(\widehat{BAD}+\widehat{DAC}=\widehat{BAC}\)
\(\Leftrightarrow\widehat{DAC}+30^0=120^0\)
hay \(\widehat{DAC}=90^0\)
b) Ta có: \(\widehat{CAE}=\dfrac{\widehat{DAC}}{2}=\dfrac{90^0}{2}=45^0\)
\(\Leftrightarrow\widehat{EAB}=120^0-45^0=75^0\)
Bài 5:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}=\dfrac{49}{50}\)