90 points please help!!!!!!!!

[tex]sin\alpha = - 0.8[/tex]Step-by-step explanation:We have , [tex]cos\alpha = \frac{3}{5}[/tex] , where [tex]\alpha[/tex] is located in IV quadrant! Let's find out value of [tex]sin\alpha[/tex] :[tex]sin\alpha = \sqrt{1-(cos\alpha )^2[/tex]⇒ [tex]sin\alpha = \sqrt{1-(cos\alpha )^2[/tex]⇒ [tex]sin\alpha = \sqrt{1-(\frac{3}{5} )^2[/tex]⇒ [tex]sin\alpha = \sqrt{1-(\frac{9}{25} )[/tex]⇒ [tex]sin\alpha = \sqrt{(\frac{25-9}{25} )[/tex]⇒ [tex]sin\alpha = \sqrt{(\frac{16}{25} )[/tex]⇒ [tex]sin\alpha = \pm {(\frac{4}{5} )[/tex]Value of [tex]sin\alpha[/tex] is dependent on which quadrant it is . Since, in question it's given that [tex]\alpha[/tex] is located in IV quadrant , So [tex]sin\alpha[/tex] is negative i.e. ⇒ [tex]sin\alpha = - {(\frac{4}{5} )[/tex]⇒ [tex]sin\alpha = - 0.8[/tex]Therefore, [tex]sin\alpha = - 0.8[/tex] .