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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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Answer:

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1) A) Name of triangle: ABO

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3) A) Name of triangle: DOC

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C) Remote Interior Angles: For ODH, it’s DOC and COD.  For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO.  For OCG, it has CDO and DOC

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Step-by-step explanation:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters.  If your teacher just wants one, it would be the middle one.  (I assume they’ll want all three, since the letters just name points, not angles.)

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B) Exterior Angles:EAC

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3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD.  For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO.  For OCG, it has CDO and DOC

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