Question

Hi, can you help me to solve this exercise please!!!

Answer 1

We are given the trigonometric ratio

`\cos (\theta)=\frac{\sqrt[]{170}}{10}`

Thus

We can represent the ratio pictorially using a right-angle triangle

We will now have to find the value of x relationship

`\begin{gathered} \text{where} \\ \text{hypotenuse}=\sqrt[]{170} \\ \text{opposite}=10 \end{gathered}`

`\cos \sec x=(1)/(\sin x)=(hypotenuse)/(opposite)=\frac{\sqrt[]{170}}{10}`

Thus, re-writing, we will have

`\sin x=\frac{10}{\sqrt[]{170}}`

Thus

`\sin \theta=\frac{10}{\sqrt[]{170}}`

In rationalized form

`\begin{gathered} \sin \theta=\frac{10}{\sqrt[]{170}}*\frac{\sqrt[]{170}}{\sqrt[]{170}} \\ \\ \sin \theta=\frac{10\sqrt[]{170}}{170}=\frac{\sqrt[]{170}}{17} \end{gathered}`

Thus

`\sin \theta=\frac{\sqrt[]{170}}{17}`