Chapter 12.6, Problem 38HP

To determine

To find: The angle $\theta$ formed by the positive x-axis and the ray.

Answer to Problem 38HP

  $\theta =-{60}^0$

Explanation of Solution

Given information:

End point of the ray is

  $P\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

Given

  $P\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

We can say that the point lies in Quadrant-IV We know any point in the coordinate plane is represented as $P\left(\cos \theta ,\sin \theta \right)$

Therefore,

  $P\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

  $=$

  $P\left(\cos \theta ,\sin \theta \right)$

On equating both

We get

  $\sin \theta =-\frac{\sqrt{3}}{2}$

  $\begin{array}{l}\sin \theta =-\frac{\sqrt{3}}{2}\\ \Rightarrow \theta ={\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\\ \Rightarrow \theta =-{60}^0\end{array}$

Therefore,

In Quadrant-IV , cosines will have positive values and sines will have negative values.

So, the angle $\theta =-{60}^0$