Need Some Help

Question

0

777

1

Choose the correct answer

Guest Nov 12, 2018

+343

+1

Best Answer

$sin^2θ +cos^2θ =1$

so $sinθ =\sqrt{1-cos^2θ}=\sqrt{\frac{36}{36}-\frac{1}{36}}=\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{6}$

so $tanθ=\frac{sinθ}{cosθ}= \frac{{\frac{\sqrt{35}}{6}}}{\frac{-1}{6}}=-\sqrt{35}$

Hope this helps!

Dimitristhym Nov 12, 2018

Dimitristhym · Answer 1 · 2018-11-12T09:16:19

$sin^2θ +cos^2θ =1$

so $sinθ =\sqrt{1-cos^2θ}=\sqrt{\frac{36}{36}-\frac{1}{36}}=\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{6}$

so $tanθ=\frac{sinθ}{cosθ}= \frac{{\frac{\sqrt{35}}{6}}}{\frac{-1}{6}}=-\sqrt{35}$

Hope this helps!

​ Need Some Help