Question

Cho sin  = 1/3 với 90\(^o\)<\(\alpha\)<180\(^o\). Tính cos \(\alpha\) và tan (180\(^o\) - \(\alpha\))

 

Answer 1

\(90^0< a< 180^0\)

=>\(cosa< 0\)

\(sin^2a+cos^2a=1\)

=>\(cos^2a=1-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}\)

mà cosa<0

nên \(cosa=-\dfrac{2\sqrt{2}}{3}\)

\(tan\left(180^0-a\right)=-tana=-\dfrac{sina}{cosa}\)

\(=-\dfrac{1}{3}:\dfrac{-2\sqrt{2}}{3}=\dfrac{1}{2\sqrt{2}}=\dfrac{\sqrt{2}}{4}\)