Answer:
The value of cos(theta) is;
[tex]\cos \theta=\frac{\sqrt[]{5}}{5}[/tex]
Explanation:
Given that;
[tex]\tan \theta=\frac{2\sqrt[]{11}}{\sqrt[]{11}}[/tex]
Recall that from trigonometry;
[tex]\tan \theta=\frac{opposite}{adjacent}[/tex]
So;
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent}=\frac{2\sqrt[]{11}}{\sqrt[]{11}} \\ \text{opposite}=2\sqrt[]{11} \\ \text{adjacent}=\sqrt[]{11} \end{gathered}[/tex]
To get the hypotenuse;
[tex]\begin{gathered} c^2=a^2+b^2 \\ c^2=(2\sqrt[]{11})^2+(\sqrt[]{11})^2 \\ c^2=44+11 \\ c^2=55 \\ c=\sqrt[]{55} \end{gathered}[/tex]
Also;
[tex]\begin{gathered} \cos \theta=\frac{adjacent}{hypotenuse}=\frac{\sqrt[]{11}}{\sqrt[]{55}}=\frac{1}{\sqrt[]{5}} \\ \cos \theta=\frac{1}{\sqrt[]{5}}\times\frac{\sqrt[]{5}}{\sqrt[]{5}} \\ \cos \theta=\frac{\sqrt[]{5}}{5} \end{gathered}[/tex]
Therefore, the value of cos(theta) is;
[tex]\cos \theta=\frac{\sqrt[]{5}}{5}[/tex]
