A point \(V\) is located on the unit circle at angle 240\(^{\circ}\) from the positive \(x\)-axis.
- Determine the quadrant in which \(V\) lies. (1 mark)
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- Find the coordinates of point \(V\). (2 marks)
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Show Worked Solution
a.
\(240^{\circ}\ \text{lies in Quadrant III}\ \ (180^{\circ} \lt 240^{\circ} \lt 270^{\circ}) \)
b. \(\text{Reference angle}\ (\theta):\ 180^{\circ}+\theta=240^{\circ}\ \ \Rightarrow \ \ \theta=60^{\circ}\)
\(x\text{-coordinate of}\ V = -\cos\,60^{\circ} = -\dfrac{1}{2}\)
\(y\text{-coordinate of}\ V = -\sin\,60^{\circ} = -\dfrac{\sqrt{3}}{2}\)
\(V\left(-\dfrac{1}{2}, -\dfrac{\sqrt{3}}{2}\right) \)