Using Sine of angles to solve for x
[tex]sin\theta=\frac{Opposite}{Hypotenuse}[/tex]
Given:
[tex]\begin{gathered} \theta=45^0 \\ Opposite=x \\ Hypotenuse=9 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} sin45^0=\frac{x}{9} \\ x=9sin45^0=\frac{9\sqrt{2}}{2} \end{gathered}[/tex]
Hence,
[tex]x=\frac{9\sqrt{2}}{2}\text{ \lparen Option 2\rparen}[/tex]
