For a $30^\circ$-$60^\circ$-$90^\circ$ triangle, recall the fixed ratio among the short leg, long leg, and hypotenuse. Think about which side is opposite each angle.

The hypotenuse is given as 10. In a $30^\circ$-$60^\circ$-$90^\circ$ triangle, how does the hypotenuse compare to the short leg, and how does the long leg compare to the short leg? Use these relationships to find $AC$.

Because $AC$ is adjacent to the $30^\circ$ angle at $A$ and $AB$ is the hypotenuse, you can use $\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AC}{10}$. In Desmos, make sure angles are in degrees, then type 10*cos(30).

Look at the numerical value that Desmos gives for 10*cos(30). Compare that decimal to the answer choices by evaluating each option (for example, type 5*sqrt(3)) and see which choice matches the result from 10*cos(30).

You are told that $\angle C = 90^\circ$ and $\angle A = 30^\circ$. The third angle $\angle B$ must be $60^\circ$ because the angles in a triangle add to $180^\circ$.

So $\triangle ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ right triangle.

In any $30^\circ$-$60^\circ$-$90^\circ$ triangle, the side lengths follow this ratio (measured relative to the angles):

• Side opposite $30^\circ$: $1$ (short leg)
• Side opposite $60^\circ$: $\sqrt{3}$ (long leg)
• Side opposite $90^\circ$: $2$ (hypotenuse)

So the hypotenuse is twice the short leg, and the long leg is $\sqrt{3}$ times the short leg.

In $\triangle ABC$:

• The hypotenuse is $AB = 10$ (opposite the right angle at $C$).
• The side opposite the $30^\circ$ angle at $A$ is $BC$ (this is the short leg).
• The side opposite the $60^\circ$ angle at $B$ is $AC$ (this is the long leg we want).

From the 30-60-90 ratio, the hypotenuse is $2 \times$ (short leg). So the short leg $BC$ must be half of the hypotenuse.

First find the short leg $BC$:

Now use the 30-60-90 ratio to find the long leg $AC$. The long leg is $\sqrt{3}$ times the short leg:

So the length of $AC$ is $5\sqrt{3}$, which corresponds to answer choice C.