Right Triangles and Trigonometry

Right triangles and trigonometry show up on roughly 3–5 questions per Digital SAT. These questions test whether you can apply the Pythagorean theorem, recognize special right triangles, and use sine, cosine, and tangent to connect angles and sides. The good news: the underlying concepts are straightforward, and many problems reward pattern recognition over heavy calculation. Knowing a few key ratios and relationships cold will save you significant time.

The Pythagorean Theorem

The Pythagorean theorem applies to every right triangle:

$a^2 + b^2 = c^2$

Here $a$ and $b$ are the two legs, and $c$ is the hypotenuse (the side opposite the 90° angle, always the longest side).

You'll use this to find a missing side when you know the other two. The SAT loves to use Pythagorean triples, which are sets of whole numbers that satisfy the theorem. Memorize these and their multiples:

• 3-4-5 (multiples: 6-8-10, 9-12-15, 15-20-25)
• 5-12-13 (multiples: 10-24-26)
• 8-15-17
• 7-24-25

Recognizing a triple lets you skip computation entirely.

Worked Example 1 (Straightforward)

In a right triangle, one leg has length 8 and the hypotenuse has length 17. What is the length of the other leg?

Set up the equation with the hypotenuse on its own side:

$8^2 + b^2 = 17^2$

$64 + b^2 = 289$

$b^2 = 225$

$b = 15$

If you recognized the 8-15-17 triple, you could write 15 immediately.

Worked Example 2 (Context Problem)

A rectangular garden measures 9 feet by 12 feet. A path runs diagonally from one corner to the opposite corner. How long is the path?

The diagonal of a rectangle creates two right triangles. The legs are 9 and 12:

$9^2 + 12^2 = c^2$

$81 + 144 = c^2$

$c^2 = 225$

$c = 15$

This is a 3-4-5 triple scaled by 3 (9-12-15).

Common trap: Forgetting which value is the hypotenuse. The hypotenuse is always opposite the right angle and always the largest number. If a problem gives you the hypotenuse and one leg, you subtract the squares rather than add them.

Special Right Triangles

Two right triangles have fixed side ratios that the SAT tests repeatedly. When you spot these angle combinations, use the ratios instead of calculating.

The 45-45-90 Triangle

Angles: 45°, 45°, 90°. Side ratio:

$1 : 1 : \sqrt{2}$

• Both legs are equal: $x$
• Hypotenuse: $x\sqrt{2}$

This triangle appears whenever you cut a square along its diagonal.

Example: If the hypotenuse of a 45-45-90 triangle is 10, find each leg.

Since the hypotenuse equals $x\sqrt{2}$:

$x\sqrt{2} = 10$

$x = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$

Each leg is $5\sqrt{2}$.

The 30-60-90 Triangle

Angles: 30°, 60°, 90°. Side ratio:

$1 : \sqrt{3} : 2$

• Side opposite 30°: $x$ (shortest)
• Side opposite 60°: $x\sqrt{3}$
• Side opposite 90° (hypotenuse): $2x$

This triangle shows up when an equilateral triangle is split in half by an altitude.

Example: In a 30-60-90 triangle, the side opposite the 60° angle is $6\sqrt{3}$. Find the hypotenuse.

The side opposite 60° equals $x\sqrt{3}$:

$x\sqrt{3} = 6\sqrt{3}$

$x = 6$

The hypotenuse is $2x = 12$.

Common trap with special right triangles: Mixing up which side is opposite which angle. The shortest side is always opposite the smallest angle (30°), and the hypotenuse is always opposite 90°. In a 45-45-90, the two legs are equal because the two acute angles are equal.

Sine, Cosine, and Tangent (SOH-CAH-TOA)

Right triangle trigonometry defines three ratios based on the position of sides relative to a chosen acute angle $\theta$:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

The mnemonic SOH-CAH-TOA encodes all three. "Opposite" and "adjacent" are always defined relative to the specific angle you're looking at, not the right angle.

Choosing the Right Ratio

When solving for a missing side:

1. Label the sides relative to the given angle: opposite, adjacent, hypotenuse.
2. Identify which side you know and which you need.
3. Pick the trig function that uses exactly those two sides.

Worked Example 1

A ladder 20 feet long leans against a wall, making a 65° angle with the ground. How high up the wall does the ladder reach?

The ladder is the hypotenuse (20 ft). The height on the wall is opposite the 65° angle. You need opposite and have hypotenuse, so use sine:

$\sin 65° = \frac{h}{20}$

$h = 20 \cdot \sin 65°$

$h \approx 20 \cdot 0.906 = 18.1 \text{ feet}$

Worked Example 2

In a right triangle, the leg opposite angle $\theta$ is 5 and the hypotenuse is 13. What is $\cos \theta$?

First, find the adjacent leg using the Pythagorean theorem:

$5^2 + a^2 = 13^2$

$a^2 = 169 - 25 = 144$

$a = 12$

Now apply the cosine definition:

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$

Notice this uses the 5-12-13 triple. The SAT often combines the Pythagorean theorem with trigonometry in a single problem.

Common trap: Confusing sine and cosine by grabbing the wrong side. If a question asks for $\sin \theta$ and you accidentally compute $\cos \theta$, your wrong answer will likely appear among the choices. Always double-check which side is opposite and which is adjacent to your specific angle.

Similarity and Trigonometric Ratios

Here's a concept the SAT tests that many students overlook: trigonometric ratios depend only on the angle, not on the size of the triangle. If two right triangles share the same acute angle, they're similar, and their side ratios are identical. That means $\sin 40°$ is the same number whether the triangle has a hypotenuse of 5 or 500.

This is why similarity matters for trigonometry. When a problem gives you two similar right triangles with different side lengths, the sine, cosine, and tangent values for corresponding angles are equal across both triangles. You can calculate a trig ratio from whichever triangle gives you enough information, then apply it to the other.

Example: Triangle ABC and triangle DEF are both right triangles with a right angle at C and F respectively. Angle A equals angle D. In triangle ABC, the side opposite angle A is 3 and the hypotenuse is 5. In triangle DEF, the hypotenuse is 15. What is the side opposite angle D?

Since the triangles are similar and share the same angle:

$\sin A = \sin D = \frac{3}{5}$

Apply this to triangle DEF:

$\frac{\text{opposite}}{15} = \frac{3}{5}$

$\text{opposite} = 15 \cdot \frac{3}{5} = 9$

Complementary Angles and the Sine-Cosine Relationship

In any right triangle, the two acute angles add up to 90°, making them complementary angles. This creates an important relationship:

$\sin \theta = \cos(90° - \theta)$

$\cos \theta = \sin(90° - \theta)$

Why? Because when you switch from one acute angle to the other, the "opposite" side becomes the "adjacent" side and vice versa. The hypotenuse stays the same. So the sine of one angle equals the cosine of the other.

Example: If $\sin 35° = 0.574$, what is $\cos 55°$?

Since $35° + 55° = 90°$, these are complementary angles:

$\cos 55° = \sin 35° = 0.574$

The SAT tests this in a slightly trickier form too:

If $\sin(3x)° = \cos(2x + 5)°$, what is $x$?

For sine to equal cosine, the angles must be complementary:

$3x + (2x + 5) = 90$

$5x + 5 = 90$

$5x = 85$

$x = 17$

What to Watch For on Test Day

1. Recognize Pythagorean triples before calculating. If you see legs of 6 and 8, the hypotenuse is 10. Spotting these patterns saves time and avoids arithmetic errors.

2. Use special right triangles whenever you see 30°, 45°, or 60°. The fixed ratios for 30-60-90 and 45-45-90 triangles are faster and more exact than using your calculator for trig.

3. Label sides relative to the angle in question. The most common mistake on trigonometry problems is swapping opposite and adjacent. Before writing any equation, clearly identify which side is which from the perspective of your specific angle.

4. Remember the complementary angle relationship. When a problem states that $\sin a° = \cos b°$, set $a + b = 90$ and solve. This shows up as its own question type and is a quick win once you know the pattern.

5. Watch for multi-step problems. The SAT often requires you to find a missing side with the Pythagorean theorem first, then compute a trig ratio, or vice versa. Read the full question before starting so you know what the final answer actually asks for.