Question 63·Easy·Geometry and Trigonometry

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In right triangle $\triangle ABC$ , $\angle C = 90^\circ$ and $\angle A = 30^\circ$ . If the hypotenuse $AB$ has length $10$ , what is the length of $AC$ ?

For right-triangle problems on the SAT, first check if the angles form a special triangle (30-60-90 or 45-45-90). If they do, use the known side ratios to move quickly between the hypotenuse and legs instead of setting up full trigonometric equations. Always map each angle to the side opposite it, identify which leg (short or long) you are asked for, and then apply the fixed ratios to avoid unnecessary calculation.

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Notice the angle measures

You are given a right angle and a $30^\circ$ angle. What does that tell you about the third angle, and what kind of special right triangle does this create?

Use the special triangle ratio

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.

Relate the hypotenuse to the legs

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 $AC$ .

Desmos Guide

Set up the trigonometric expression

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).

Interpret the result

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).

Step-by-step Explanation

Identify the type of triangle

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.

Recall the special 30-60-90 side ratio

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.

Match the sides of the ratio to the given triangle

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.

Compute the short leg, then the long leg (AC)

First find the short leg $BC$ :

BC = \frac{AB}{2} = \frac{10}{2} = 5

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

AC = 5 \cdot \sqrt{3} = 5\sqrt{3}

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

Question 63·Easy·Geometry and Trigonometry

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation