[tex]if\ \sin\Theta \ \textgreater \ 0\ and\ \tan\Theta \ \textless \ 0\ then\ \Theta\in\underbrace{(90^o;\ 180^o)}_{\left(\dfrac{\pi}{2};\ \pi\right)}\\\\therefore\ \cos\Theta \ \textless \ 0.[/tex]
[tex]\text{Use:}\ \sin^2x+\cos^2x=1\\\\\left(\dfrac{3}{7}\right)^2+\cos^2\Theta=1\\\\\dfrac{9}{49}+\cos^2\Theta=1\ \ \ |-\dfrac{9}{49}\\\\\cos^2\Theta=\dfrac{40}{49}\to\cos\Theta=-\sqrt{\dfrac{40}{49}}\\\\\cos\Theta=-\dfrac{\sqrt{40}}{\sqrt{49}}\\\\\cos\Theta=-\dfrac{\sqrt{4\cdot10}}{7}\\\\\boxed{\cos\Theta=-\dfrac{2\sqrt{10}}{7}}[/tex]
