Question

Help me please i want to know if i’m doing it right

Answer 1

The value of cos(theta) is;

`\cos \theta=\frac{\sqrt[]{5}}{5}`

Step-by-step explanation:

Given that;

`\tan \theta=\frac{2\sqrt[]{11}}{\sqrt[]{11}}`

Recall that from trigonometry;

`\tan \theta=(opposite)/(adjacent)`

So;

`\begin{gathered} \tan \theta=(opposite)/(adjacent)=\frac{2\sqrt[]{11}}{\sqrt[]{11}} \\ \text{opposite}=2\sqrt[]{11} \\ \text{adjacent}=\sqrt[]{11} \end{gathered}`

To get the hypotenuse;

`\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}`

Also;

`\begin{gathered} \cos \theta=(adjacent)/(hypotenuse)=\frac{\sqrt[]{11}}{\sqrt[]{55}}=\frac{1}{\sqrt[]{5}} \\ \cos \theta=\frac{1}{\sqrt[]{5}}*\frac{\sqrt[]{5}}{\sqrt[]{5}} \\ \cos \theta=\frac{\sqrt[]{5}}{5} \end{gathered}`

Therefore, the value of cos(theta) is;

`\cos \theta=\frac{\sqrt[]{5}}{5}`