WinnerV644658 WinnerV644658
  • 30-10-2022
  • Mathematics
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find it's a y 30° 9 square root of 3

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TamarahT454693 TamarahT454693
  • 30-10-2022

Let's use sin and cosine identities:

[tex]\begin{gathered} \cos (\theta)=\frac{adjacent}{\text{hypotenuse}} \\ \\ \cos (30)=\frac{9\sqrt[]{3}}{y} \\ \\ \text{Solving for y:} \\ y=\frac{9\sqrt[]{3}}{\cos (30)}=18 \\ \end{gathered}[/tex][tex]\begin{gathered} \sin (\theta)=\frac{opposite}{\text{hypotenuse}} \\ \\ \sin (30)=\frac{x}{18} \\ \text{Solving for x:} \\ x=18\cdot\sin (30)=9 \end{gathered}[/tex]

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