Question

Дайте відповідь на питання на фото

Answer 1

Ответ:

1) $\frac{\sqrt{10}}{10}$

2) $\frac{5}{12}$

Объяснение:

1)

$sin\alpha\\\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\\\\frac{sin\alpha}{cos\alpha}=3\\\\sin\alpha=3 cos\alpha$

$cos\alpha\\\\sin^2\alpha+cos^2\alpha=1\\\\\left( 3 cos\alpha\right)^2+cos^2\alpha =1\\\\9cos^2\alpha+cos^2\alpha=1\\\\ 10cos^2\alpha=1\ \ \ |:10 \\\\cos^2\alpha= \frac{ 1 }{ 10 } \\\\cos\alpha=\sqrt{ \frac{ 1 }{ 10 } } \\\\cos\alpha= \frac{\sqrt{10}}{10}$

2)

$cos\alpha\\\\sin^2\alpha+cos^2\alpha=1\\\\\left( \frac{ 12 }{ 13 } \right)^2 +cos^2\alpha=1\\\\ \frac{ 144 }{ 169 } +cos^2x=1\\\\cos^2\alpha=1- \frac{ 144 }{ 169 } \\\\cos^2\alpha= \frac{ 25 }{ 169 } \\\\cos\alpha=\sqrt{ \frac{ 25 }{ 169 } }\\\\cos\alpha=\frac{5}{13}$

$ctg\alpha\\\\ctg\alpha=\frac{sin\alpha}{cos\alpha}\\\\ctg\alpha=\frac{\frac{12}{13}}{\frac{5}{13}}\\\\ctg\alpha=\frac{5}{12}$