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Alex73 [517]
3 years ago
10

Please help me I will fail the exam help me please

Mathematics
1 answer:
Sergio039 [100]3 years ago
5 0
The picture is not showing up :(
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If tan theta = 10/13 and cos theta > 0 then sin2theta is what ?
svetoff [14.1K]

sin2\alpha  = \frac{260}{269}

<u>Step-by-step explanation:</u>

We have , Tan\alpha  = \frac{Perpendicular}{Base} = \frac{10}{13},

We know that sin\alpha  = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }

Substituting values of P & B , sin\alpha  = \frac{10}{\sqrt{10^{2} + 13^{2}} }  = \frac{10}{\sqrt{269} }

Now , sin2\alpha  = 2sin\alpha cos\alpha  = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }

⇒sin2\alpha  = \frac{10}{\sqrt{269} } ×\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }×2

⇒ sin2\alpha  = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} }  )

⇒sin2\alpha  = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} }  )

⇒sin2\alpha  = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} }  )

⇒sin2\alpha  = \frac{260}{269}

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9/20

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Step-by-step explanation:

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