Question

найдите tga,если cos a=5/13 и 0

Answer 1

Ответ:

$cosa=\dfrac{5}{13}\\\\\\\displaystyle sin^2a=1-cos^2a=1-\frac{25}{169}=\frac{144}{169}\ \ ,\ \ sina=\pm \frac{12}{13}\\\\\\0<a<\dfrac{\pi }{2}\ \ \ \Rightarrow \ \ \ sina>0\ \ ,\ \ sina=\frac{12}{13}\\\\\\tga=\frac{sina}{cosa}=\frac{12/13}{5/13}=\frac{12}{5}$