Respuesta :
[tex]sin2\alpha = \frac{260}{269}[/tex]
Step-by-step explanation:
We have , [tex]Tan\alpha = \frac{Perpendicular}{Base} = \frac{10}{13}[/tex],
We know that [tex]sin\alpha = \frac{Perpendicular}{Hypotenuse} = \frac{Perpendicular}{\sqrt[2]{(Perpendicualr)^{2} + (Base)^{2})} }[/tex]
Substituting values of P & B , [tex]sin\alpha = \frac{10}{\sqrt{10^{2} + 13^{2}} } = \frac{10}{\sqrt{269} }[/tex]
Now , [tex]sin2\alpha = 2sin\alpha cos\alpha = 2sin\alpha \sqrt{1 - (sin\alpha)^{2} }[/tex]
⇒[tex]sin2\alpha =[/tex] [tex]\frac{10}{\sqrt{269} }[/tex] ×[tex]\sqrt{1 - (\frac{10}{\sqrt{269} })^{2} }[/tex]×2
⇒ [tex]sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{269 - 100}{269} } )[/tex]
⇒[tex]sin2\alpha = \frac{20}{\sqrt{269} }( \sqrt{\frac{169 }{269} } )[/tex]
⇒[tex]sin2\alpha = \frac{260}{\sqrt{269} }( \sqrt{\frac{1}{269} } )[/tex]
⇒[tex]sin2\alpha = \frac{260}{269}[/tex]
