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

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

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

Answer

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:

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