Question

знайдіть cosА i tgА, якщо sinА =0,8; пи/2​<A<пи

Answer 1

$\displaystyle\bf\\\frac{\pi }{2} < A < \pi \ \ \ \Rightarrow \ \ \ CosA < 0 \ \ ; \ \ tgA < 0\\\\SinA=0,8\\\\CosA=-\sqrt{1-Sin^{2} A} =-\sqrt{1-(0,8)^{2} } =-\sqrt{1-0,64}=\\\\=-\sqrt{0,36}=-0,6\\\\\\ tgA=\frac{SinA}{CosA} =\frac{0,8}{-0,6}=-\frac{4}{3} =-1\frac{1}{3}$