a) Áp dụng hệ thức:
\(sin^2\alpha+cos^2\alpha=1\)
<=>\(sin^2\alpha+\left(\dfrac{5}{13}\right)^2=1\)
<=>\(sin^2\alpha+\dfrac{25}{169}=1\)
<=>\(sin^2\alpha=1-\dfrac{25}{169}=\dfrac{144}{169}\)
<=>\(sin\alpha=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\)
Ta có: \(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}=\dfrac{12}{13}.\dfrac{13}{5}=\dfrac{12}{5}\)