### Instruction: Compute $\sin 60^\circ$.### Response: I recognize that 60 degrees is a special angle that appears in an equilateral triangle, which has all sides equal and all angles equal to 60 degrees.
I can draw an equilateral triangle and label its side length as s. Then I can bisect one of the angles and draw the altitude to the opposite side, forming a right triangle.
In this right triangle, the hypotenuse is s, the base is s/2, and the height is unknown.
I can use the Pythagorean theorem to find the height: $(s/2)^2 + h^2 = s^2$.
Simplifying this equation, I get $h^2 = s^2 - s^2/4 = 3s^2/4$.
Taking the positive square root, I get $h = \sqrt{3}s/2$.
Now I can use the definition of sine as the ratio of the opposite side to the hypotenuse.