Question

use the special triangles on the unit circle to determine $\theta$ in degrees when $sin\theta = \frac{sqrt{3}}{2}$. (1 point)$\bigcirc$ $90^circLXB0\bigcirc$ $60^circ$$\bigcirc$ $45^circ$

Explanation:

Step1: Recall 30-60-90 triangle ratios

In a 30-60-90 right triangle, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$. For $\theta=60^\circ$, opposite side is $\frac{\sqrt{3}}{2}$ (unit hypotenuse).

Step2: Match with given sine value

Given $\sin\theta = \frac{\sqrt{3}}{2}$, this matches $\sin60^\circ$.

Answer:

C. 60°