KyleQ452598 KyleQ452598
  • 26-10-2022
  • Mathematics
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Geometry or algebra 2 thanks so much its asking for cot

Geometry or algebra 2 thanks so much its asking for cot class=

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AjourneeI30272 AjourneeI30272
  • 26-10-2022

Answer: We have to evaluate the cotangent for the given x and y coordinate points:

[tex]\begin{gathered} (x,y)=(\frac{\sqrt{3}}{4},-\frac{\sqrt{13}}{4}) \\ \\ x=\frac{\sqrt{3}}{4} \\ \\ y=-\frac{\sqrt{13}}{4} \end{gathered}[/tex]

The value of the cot function is:

[tex]\begin{gathered} cot(\theta)=\frac{x}{y} \\ \\ \therefore\rightarrow \\ \\ =\frac{\sqrt{3}}{4}\div-\frac{\sqrt{13}}{4}=-\frac{\sqrt{3}}{\sqrt{13}}=-\sqrt{\frac{3}{13}} \\ \\ =-\sqrt{\frac{3}{13}} \end{gathered}[/tex]

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