Question

try it
what are the input and output values for determining
the sine of 60°?
○ input: \\(\frac{2}{\sqrt{3}}\\); output: 60°
○ input: 60°; output: \\(\frac{\sqrt{3}}{2}\\)
○ input: 60°; output: \\(\frac{2}{\sqrt{3}}\\)
○ input: \\(\frac{\sqrt{3}}{2}\\); output: 60°
(image of a right triangle with vertices a, b, c, right-angled at c. angle at a is 30°, angle at b is 60°. side ac is \\(8\sqrt{3}\\), side bc is 8, side ab is 16.)

Explanation:

Step1: Recall Sine Definition

Sine of an angle in a right triangle is $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. For $\theta = 60^\circ$, the input is the angle $60^\circ$.

Step2: Identify Opposite and Hypotenuse

In $\triangle ABC$, for $\angle B = 60^\circ$ (wait, no, let's check the triangle. Wait, angle at B? Wait, angle at A is $30^\circ$, right angle at C. So for angle at B: wait, no, let's see the sides. Wait, hypotenuse is AB = 16, opposite side to $60^\circ$ (let's find which angle is $60^\circ$. Wait, angle at A is $30^\circ$, so angle at B is $60^\circ$ (since triangle angles sum to $180^\circ$: $30 + 90 + 60 = 180$). So for angle $60^\circ$ (angle at B), the opposite side is AC = $8\sqrt{3}$, hypotenuse is AB = 16? Wait, no, wait: in right triangle, sine of an angle is opposite over hypotenuse. Wait, maybe better to recall the special triangle: 30-60-90 triangle has sides in ratio $1 : \sqrt{3} : 2$. So for $60^\circ$, $\sin(60^\circ)=\frac{\sqrt{3}}{2}$. So the input is the angle $60^\circ$, output is $\frac{\sqrt{3}}{2}$. Let's check the options:

Option 2: input: $60^\circ$; output: $\frac{\sqrt{3}}{2}$ — this matches. Let's verify:

$\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}$. In the triangle, for angle $60^\circ$ (angle at B), opposite side is AC = $8\sqrt{3}$, hypotenuse is AB = 16. So $\sin(60^\circ)=\frac{8\sqrt{3}}{16}=\frac{\sqrt{3}}{2}$. Yes, that's correct.

Answer:

B. input: $60^\circ$; output: $\frac{\sqrt{3}}{2}$