Question

when finding (cos \theta = \frac{sqrt{3}}{2}) on a unit circle, what type of special right triangle or a ray is used? (1 point) (\bigcirc) 30-60-90 triangle (\bigcirc) 45-45-90 triangle (\bigcirc) a vertical ray from the origin (\bigcirc) a horizontal ray from the origin

Explanation:

1. Recall the properties of special right triangles and the unit circle:

• For a 30 - 60 - 90 triangle, the ratios of the sides are \(1:\sqrt{3}:2\). In the unit circle, \(\cos\theta=\frac{x}{r}\), where \(r = 1\) (unit circle). If \(\cos\theta=\frac{\sqrt{3}}{2}\), then \(x=\frac{\sqrt{3}}{2}\) and \(r = 1\). Using the Pythagorean theorem \(y=\sqrt{r^{2}-x^{2}}=\sqrt{1 - (\frac{\sqrt{3}}{2})^{2}}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\). The angle \(\theta\) corresponding to \(x = \frac{\sqrt{3}}{2}\) and \(y=\frac{1}{2}\) is \(30^{\circ}\) or \(60^{\circ}\) (since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\) and \(\cos60^{\circ}=\frac{1}{2}\), here we consider the angle with \(\cos\theta=\frac{\sqrt{3}}{2}\) which is \(30^{\circ}\) from the x - axis, and the triangle formed is a 30 - 60 - 90 triangle).
• For a 45 - 45 - 90 triangle, the cosine of the non - right angles is \(\frac{\sqrt{2}}{2}\), not \(\frac{\sqrt{3}}{2}\), so this option is incorrect.
• A vertical ray from the origin would have \(x = 0\), so \(\cos\theta=0\), not \(\frac{\sqrt{3}}{2}\), so this option is incorrect.
• A horizontal ray from the origin would have \(y = 0\), and \(\cos\theta = 1\) (when \(\theta=0^{\circ}\)) or \(\cos\theta=- 1\) (when \(\theta = 180^{\circ}\)), not \(\frac{\sqrt{3}}{2}\), so this option is incorrect.

Answer:

A. 30 - 60 - 90 triangle