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Arte-miy333 [17]
3 years ago
9

If sin Θ = 5/6 , what are the values of cos Θ and tan Θ?

Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0
For the answer to the question above, I'll provide a solution to my answer below
sin²θ + cos²θ = 1 

<span>so, cos²θ = 1 - (5/6)² => 11/36 </span>

<span>=> cosθ = ±√11/6 </span>

<span>Also, tanθ = sinθ/cosθ => (5/6)/(±√11/6) </span>

<span>i.e. tanθ = ±5/√11 
</span>I hope my answer helped you. Feel free to ask more questions. Have a nice day!
kirza4 [7]3 years ago
6 0

Answer:  The required values are

\cos\theta=\pm\dfrac{\sqrt{11}}{6},\\\\\\\tan\theta=\pm\dfrac{5\sqrt{11}}{11}.

Step-by-step explanation:  We are given the following value :

\sin\theta=\dfrac{5}{6}

We are to find the values of \cos\theta and \tan\theta.

We will be using the following formulas :

(i)\sin^2x+\cos^2x=1,\\\\\\(ii)\tan x=\dfrac{\sin x}{\cos x}.

We have

\cos\theta\\\\\\=\pm\sqrt{1-\sin^2\theta}\\\\\\=\pm\sqrt{1-\dfrac{25}{36}}\\\\\\=\pm\sqrt{\dfrac{11}{36}}\\\\\\=\pm\dfrac{\sqrt{11}}{6}

and

\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac{5}{6}}{\pm\frac{\sqrt{11}}{6}}=\pm\dfrac{5}{\sqrt{11}}=\pm\dfrac{5\sqrt{11}}{11}.

Thus, the required values are

\cos\theta=\pm\dfrac{\sqrt{11}}{6},\\\\\\\tan\theta=\pm\dfrac{5\sqrt{11}}{11}.

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