Question

select the correct answer. the shortest side of a right triangle measures $3\sqrt{3}$ inches. one angle of the triangle measures $60^\circ$. what is the length, in inches, of the hypotenuse of the triangle? a. $6\sqrt{2}$ b. 3 c. $6\sqrt{3}$ d. 6

Explanation:

Step1: Identify triangle angles

This is a 30-60-90 right triangle: the right angle is $90^\circ$, given angle is $60^\circ$, so the third angle is $180^\circ - 90^\circ - 60^\circ = 30^\circ$. The shortest side is opposite the $30^\circ$ angle.

Step2: Recall 30-60-90 triangle ratios

In a 30-60-90 triangle, the hypotenuse is twice the length of the side opposite the $30^\circ$ angle (the shortest side). Let the shortest side be $s = 3\sqrt{3}$, hypotenuse $h = 2s$.

Step3: Calculate hypotenuse length

$h = 2 \times 3\sqrt{3} = 6\sqrt{3}$
Wait, correction: Wait, no—wait, in 30-60-90, sides are $s$ (30°), $s\sqrt{3}$ (60°), $2s$ (hypotenuse). The shortest side is $s$, but here the given shortest side is $3\sqrt{3}$, which is $s$. So hypotenuse is $2s = 2 \times 3\sqrt{3}$? No, wait no—wait, if the angle is 60°, the side adjacent to 60° is the shortest side (30° opposite). Wait, using sine: $\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\text{shortest side}}{h}$. $\sin(30^\circ) = \frac{1}{2}$, so $\frac{3\sqrt{3}}{h} = \frac{1}{2}$? No, that's wrong. Wait, no: if one angle is 60°, the side opposite 60° is $s\sqrt{3}$, the side opposite 30° is $s$ (shortest). If the shortest side is $3\sqrt{3}$, that is $s = 3\sqrt{3}$, so hypotenuse is $2s = 6\sqrt{3}$? No, wait no—wait, let's use cosine: $\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}$, adjacent to 60° is the shortest side. $\cos(60^\circ) = 0.5$, so $\frac{3\sqrt{3}}{h} = 0.5$, so $h = 6\sqrt{3}$? No, wait no, wait the options have D as 6. Wait, I mixed up: if the given side is opposite 60°? No, the problem says "the shortest side". The shortest side in right triangle is opposite the smallest angle, which is 30°. So if shortest side is $3\sqrt{3}$, that is $s$, hypotenuse is $2s = 6\sqrt{3}$? But option C is $6\sqrt{3}$. Wait, but let's check with tangent: $\tan(60^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{longer leg}}{\text{shortest leg}} = \sqrt{3}$, so longer leg is $3\sqrt{3} \times \sqrt{3} = 9$. Then hypotenuse via Pythagoras: $\sqrt{(3\sqrt{3})^2 + 9^2} = \sqrt{27 + 81} = \sqrt{108} = 6\sqrt{3}$. Yes, that's correct.

Answer:

C. $6\sqrt{3}$