Respuesta :

Answer:

[tex]\frac{\sqrt{117} }{11}[/tex]

Step-by-step explanation:

sin²β + cos²β = 1

sin²β = 1 - ( - [tex]\frac{2}{11}[/tex] )²

sin²β = 1 - [tex]\frac{4}{121}[/tex] = [tex]\frac{117}{121}[/tex]

sin β > 0 ( Q. ll )

sin β = [tex]\frac{\sqrt{117} }{11}[/tex]

msm555

sin[tex]\theta_{1}=\frac{\sqrt{117}}{11}[/tex]

Answer:

Solution given:

[tex]\bold{Cos \theta_{1}=\frac{-2}{11}}[/tex]

squaring both side we get

[tex]\bold{Cos² \theta_{1}=\frac{4}{121}}[/tex]

1-sin²[tex]\theta_{1}=\frac{4}{121}[/tex]

sin²[tex]\theta_{1}=1-\frac{4}{121}[/tex]

sin[tex]\theta_{1}=\sqrt{\frac{117}{121}}[/tex]

sin[tex]\theta_{1}=\frac{\sqrt{117}}{11}[/tex]