Đáp án:
\(\left[ \begin{array}{l}\begin{cases}\sin A=\dfrac{5}{13}\\\cos A=\dfrac{12}{13}\\\cot A=\dfrac{12}{5}\\\end{cases}\\\begin{cases}\sin A=-\dfrac{5}{13}\\\cos A=-\dfrac{12}{13}\\\cot A=\dfrac{12}{5}\\\end{cases}\end{array} \right.\)
Giải thích các bước giải:
`tanA=5/12`
`=>cotA=1/tanA=1:5/12=12/5`
`tanA=5/12`
`=>(sinA)/(cosA)=5/12`
`=>sinA=(5cosA)/12`
Mặt khác:
`sin^2A+cos^2A=1`
`<=>(25cos^2A)/144+cos^2A=1`
`<=>(169cos^2A)/144=1`
`<=>cos^2A=144/169`
`<=>` \(\left[ \begin{array}{l}\cos A=\dfrac{12}{13}\\\cos A=-\dfrac{12}{13}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}\sin A=\dfrac{5\cos A}{12}=\dfrac{5}{13}\\\sin A=\dfrac{5\cos A}{12}=-\dfrac{5}{13}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}\begin{cases}\sin A=\dfrac{5}{13}\\\cos A=\dfrac{12}{13}\\\cot A=\dfrac{12}{5}\\\end{cases}\\\begin{cases}\sin A=-\dfrac{5}{13}\\\cos A=-\dfrac{12}{13}\\\cot A=\dfrac{12}{5}\\\end{cases}\end{array} \right.\)
