Question

1. the point $(-5\sqrt{3},15)$ lies on the terminal side of an angle $\theta$ in standard position. find the value of $\theta$ in degrees.

Explanation:

Step1: Calculate the reference angle

We know that for a point \((x,y)=(-5\sqrt{3},15)\), \(\tan\theta=\frac{y}{x}=\frac{15}{- 5\sqrt{3}}=-\sqrt{3}\). The reference angle \(\theta_{r}\) is found by \(\tan\theta_{r}=\vert\frac{y}{x}\vert=\sqrt{3}\), so \(\theta_{r} = 60^{\circ}\)

Step2: Determine the quadrant

Since \(x=-5\sqrt{3}<0\) and \(y = 15>0\), the point \((-5\sqrt{3},15)\) lies in the second quadrant.

Step3: Find the angle \(\theta\)

In the second quadrant, \(\theta=180^{\circ}-\theta_{r}\). Substituting \(\theta_{r} = 60^{\circ}\), we get \(\theta=180^{\circ}-60^{\circ}=120^{\circ}\)

Answer:

\(120^{\circ}\)