The area of the shaded region
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Okay, notice one thing right off the bat? 60 degrees on a right triangle (unit circle w/ terminal arm) will create the infamous 30-60-90 triangle.
We can use this to our advantage. Since sin is the y-coordinate on the unit circle. We need to solve for the longest side of a 30-60-90 triangle.
We find that the long side (y) of our 30-60-90 triangle is sqrt(3)/2. Since this is in quadrant 1, we know it is positive.
So -- sin60° = sqrt(3)/2
