Question

let $\sin(60)=\frac{\sqrt{3}}{2}$. enter the angle measure $(\theta)$, in degrees, for $\cos(\theta)=\frac{\sqrt{3}}{2}$.

Explanation:

Step1: Recall the co - function identity

We know that \(\sin(A)=\cos(90^{\circ}-A)\). Given \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), then \(\cos(90^{\circ} - 60^{\circ})=\frac{\sqrt{3}}{2}\).

Step2: Calculate the angle

\(90^{\circ}-60^{\circ}=30^{\circ}\). Also, we can use the inverse cosine function. If \(\cos(\theta)=\frac{\sqrt{3}}{2}\), then \(\theta=\arccos(\frac{\sqrt{3}}{2})\).
Since the range of the inverse cosine function \(y = \arccos(x)\) is \(0^{\circ}\leq\theta\leq180^{\circ}\), and we know from the unit circle that \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\).

Answer:

\(30\)