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

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

$\cos \sec x=\frac{1}{\sin x}=\frac{hypotenuse}{opposite}=\frac{\sqrt[]{170}}{10}$

Thus, re-writing, we will have

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

Thus

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

In rationalized form

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

Thus

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