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# ​ Need Some Help

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Nov 12, 2018

#1
+316
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$$sin^2θ +cos^2θ =1$$

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!

Nov 12, 2018

#1
+316
+2

$$sin^2θ +cos^2θ =1$$

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