Question

срочнооооооооооооо!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Объяснение:

$sin\alpha =\frac{5}{13}\ \ \ \ 0^0< \alpha <180^0\ \ \ \ tg\alpha =?\\sin^2\alpha +cos^2\alpha =1\\cos^2\alpha =1-sin^2\alpha \\cos^2\alpha =1-(\frac{5}{13})^2=1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}\\cos\alpha =б\sqrt{\frac{144}{169} } =б\frac{12}{13}.\\tg\alpha =\frac{sin\alpha }{cos\alpha }=\frac{\frac{5}{13} }{б\frac{12}{13} }=б\frac{5}{12} .$

Ответ: 4) ±5/12.