Question

Given the following point on the unit circle, find the angle, to the nearest tenth of adegree (if necessary), of the terminal side through that point, o<=0<360.

Answer 1

Since the x-coordinate of the point is negative and the y-coordinate is positive, the point is in the second quadrant (90° < angle < 180°)

Then, in order to find the angle, we can use the relations:

`\begin{gathered} \cos (\theta)=-\frac{\sqrt[]{2}}{5} \\ \sin (\theta)=\frac{\sqrt[]{23}}{5} \end{gathered}`

Using a calculator and knowing that the angle is between 90° and 180°, we have:

`\begin{gathered} \theta=\cos ^(-1)(-\frac{\sqrt[]{2}}{5})=106.4\degree \\ \theta=\sin ^(-1)(\frac{\sqrt[]{23}}{5})=106.4\degree \end{gathered}`

So the angle is 106.4°.