Geometry and Trigonometry

Apply geometry concepts including area, volume, angles, and basic trigonometry.

🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Geometry and Trigonometry on the SAT

Essential Geometry Formulas (These ARE Given on the SAT)

The SAT provides these formulas at the beginning of each math section:

| Shape | Formula | |---|---| | Circle area | $A = \pi r^2$ | | Circle circumference | $C = 2\pi r$ | | Rectangle area | $A = lw$ | | Triangle area | $A = \frac{1}{2}bh$ | | Pythagorean theorem | $a^2 + b^2 = c^2$ | | Special right triangles | 30-60-90 and 45-45-90 | | Volume of box | $V = lwh$ | | Volume of cylinder | $V = \pi r^2 h$ | | Volume of sphere | $V = \frac{4}{3}\pi r^3$ | | Volume of cone | $V = \frac{1}{3}\pi r^2 h$ | | Volume of pyramid | $V = \frac{1}{3}Bh$ |

Angles

Angle Relationships

Complementary angles: sum = $90°$
Supplementary angles: sum = $180°$
Vertical angles: equal (across from each other at an intersection)
Triangle angle sum: $180°$

Parallel Lines Cut by a Transversal

Corresponding angles are equal
Alternate interior angles are equal
Alternate exterior angles are equal
Co-interior (same-side interior) angles are supplementary ( $180°$ )

Triangles

Key Properties

The sum of interior angles = $180°$
The longest side is opposite the largest angle
Triangle inequality: The sum of any two sides > the third side

Special Right Triangles

45-45-90: Sides in ratio $1 : 1 : \sqrt{2}$ If legs = $a$ , hypotenuse = $a\sqrt{2}$

30-60-90: Sides in ratio $1 : \sqrt{3} : 2$

Short leg (opposite 30°) = $a$
Long leg (opposite 60°) = $a\sqrt{3}$
Hypotenuse (opposite 90°) = $2a$

Similar Triangles

If two triangles are similar:

Corresponding angles are equal
Corresponding sides are proportional
Area ratio = (side ratio)²

Trigonometry

SOH CAH TOA

For a right triangle with angle $\theta$ :

$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \qquad \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \qquad \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Complementary Angle Relationship

$\sin(x) = \cos(90° - x)$ $\cos(x) = \sin(90° - x)$

This is a common SAT trick: "co" in cosine stands for "complement."

Unit Circle Values (Most Common on SAT)

| Angle | $\sin$ | $\cos$ | $\tan$ | |---|---|---|---| | $0°$ | $0$ | $1$ | $0$ | | $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ | | $45°$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ | | $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | | $90°$ | $1$ | $0$ | undefined |

Radians vs. Degrees

$180° = \pi \text{ radians}$ $\text{To convert: degrees} \times \frac{\pi}{180} = \text{radians}$ $\text{To convert: radians} \times \frac{180}{\pi} = \text{degrees}$

Coordinate Geometry

Distance Formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Midpoint Formula

$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

SAT Question Types

Type 1: Find a Missing Angle

Use angle sum properties (triangle = 180°, supplementary, parallel lines).

Type 2: Apply Area/Volume Formulas

Plug values into the given formulas and solve.

Type 3: Right Triangle Trigonometry

Set up a trig ratio and solve for the unknown side or angle.

Type 4: Special Right Triangles

Recognize 30-60-90 or 45-45-90 patterns and use ratios.

Type 5: Similar Triangles

Set up proportions from corresponding sides.

Common SAT Mistakes

Using the wrong trig ratio — label O, A, H carefully
Confusing 30-60-90 ratios — the longest leg is $a\sqrt{3}$ , not $2a$
Forgetting to use the formula page — it's provided, reference it!
Calculator in wrong mode — make sure it's in degrees (not radians) unless specified
Assuming figures are drawn to scale — they may not be!

📚 Practice Problems

1Problem 1easy

❓ Question:

In a right triangle, one leg is 6 and the hypotenuse is 10. What is the length of the other leg?

💡 Show Solution

Pythagorean Theorem: $a^2 + b^2 = c^2$

$6^2 + b^2 = 10^2$ $36 + b^2 = 100$ $b^2 = 64$ $b = 8$

Answer: The other leg is 8.

Shortcut: This is a 6-8-10 triangle (a multiple of the 3-4-5 Pythagorean triple: $3 \times 2 = 6$ , $4 \times 2 = 8$ , $5 \times 2 = 10$ ).

2Problem 2easy

❓ Question:

In a right triangle, one leg is 6 and the hypotenuse is 10. What is the length of the other leg?

💡 Show Solution

Pythagorean Theorem: $a^2 + b^2 = c^2$

$6^2 + b^2 = 10^2$ $36 + b^2 = 100$ $b^2 = 64$ $b = 8$

Answer: The other leg is 8.

Shortcut: This is a 6-8-10 triangle (a multiple of the 3-4-5 Pythagorean triple: $3 \times 2 = 6$ , $4 \times 2 = 8$ , $5 \times 2 = 10$ ).

3Problem 3medium

❓ Question:

In a 30-60-90 triangle, the side opposite the 30° angle is 5. What is the length of the hypotenuse?

💡 Show Solution

30-60-90 ratio: $1 : \sqrt{3} : 2$

The side opposite 30° is the shortest side = $a = 5$ .

The hypotenuse = $2a = 2(5) = 10$ .

(The side opposite 60° = $a\sqrt{3} = 5\sqrt{3} \approx 8.66$ )

Answer: Hypotenuse = 10

4Problem 4medium

❓ Question:

In a 30-60-90 triangle, the side opposite the 30° angle is 5. What is the length of the hypotenuse?

💡 Show Solution

30-60-90 ratio: $1 : \sqrt{3} : 2$

The side opposite 30° is the shortest side = $a = 5$ .

The hypotenuse = $2a = 2(5) = 10$ .

(The side opposite 60° = $a\sqrt{3} = 5\sqrt{3} \approx 8.66$ )

Answer: Hypotenuse = 10

5Problem 5medium

❓ Question:

In a right triangle, $\sin A = \frac{5}{13}$ . What is $\cos A$ ?

💡 Show Solution

Step 1: From $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}$ :

Opposite = 5
Hypotenuse = 13

Step 2: Find the Adjacent side using the Pythagorean theorem: $a^2 + 5^2 = 13^2$ $a^2 = 169 - 25 = 144$ $a = 12$

Step 3: $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{12}{13}$

Answer: $\cos A = \frac{12}{13}$

Shortcut: This is the 5-12-13 Pythagorean triple.

6Problem 6medium

❓ Question:

In a right triangle, $\sin A = \frac{5}{13}$ . What is $\cos A$ ?

💡 Show Solution

Step 1: From $\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}$ :

Opposite = 5
Hypotenuse = 13

Step 2: Find the Adjacent side using the Pythagorean theorem: $a^2 + 5^2 = 13^2$ $a^2 = 169 - 25 = 144$ $a = 12$

Step 3: $\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{12}{13}$

Answer: $\cos A = \frac{12}{13}$

Shortcut: This is the 5-12-13 Pythagorean triple.

7Problem 7hard

❓ Question:

Two sides of a triangle are 8 and 15 and the included angle is 60°. What is the area of the triangle?

💡 Show Solution

Formula for area with an included angle: $A = \frac{1}{2}ab\sin C$

Where $a = 8$ , $b = 15$ , and $C = 60°$ :

$A = \frac{1}{2}(8)(15)\sin 60°$ $= \frac{1}{2}(120)\left(\frac{\sqrt{3}}{2}\right)$ $= 60 \cdot \frac{\sqrt{3}}{2}$ $= 30\sqrt{3} \approx 51.96$

Answer: $30\sqrt{3}$ square units (approximately 51.96)

Note: This formula is not on the SAT formula sheet but appears in harder problems.

8Problem 8hard

❓ Question:

Two sides of a triangle are 8 and 15 and the included angle is 60°. What is the area of the triangle?

💡 Show Solution

Formula for area with an included angle: $A = \frac{1}{2}ab\sin C$

Where $a = 8$ , $b = 15$ , and $C = 60°$ :

$A = \frac{1}{2}(8)(15)\sin 60°$ $= \frac{1}{2}(120)\left(\frac{\sqrt{3}}{2}\right)$ $= 60 \cdot \frac{\sqrt{3}}{2}$ $= 30\sqrt{3} \approx 51.96$

Answer: $30\sqrt{3}$ square units (approximately 51.96)

Note: This formula is not on the SAT formula sheet but appears in harder problems.

9Problem 9expert

❓ Question:

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? How far is the base of the ladder from the wall?

💡 Show Solution

Step 1: Draw the right triangle:

Hypotenuse = ladder = 20 ft
Angle with ground = 65°
Height = opposite side
Distance from wall = adjacent side

Step 2: Find the height (opposite): $\sin 65° = \frac{\text{height}}{20}$ $\text{height} = 20 \sin 65° = 20(0.9063) \approx 18.13 \text{ ft}$

Step 3: Find the distance from wall (adjacent): $\cos 65° = \frac{\text{distance}}{20}$ $\text{distance} = 20 \cos 65° = 20(0.4226) \approx 8.45 \text{ ft}$

Check: $18.13^2 + 8.45^2 \approx 328.7 + 71.4 = 400.1 \approx 20^2$ ✓

Answer: Height ≈ 18.13 ft, Distance from wall ≈ 8.45 ft

10Problem 10expert

❓ Question:

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? How far is the base of the ladder from the wall?

💡 Show Solution

Step 1: Draw the right triangle:

Hypotenuse = ladder = 20 ft
Angle with ground = 65°
Height = opposite side
Distance from wall = adjacent side

Step 2: Find the height (opposite): $\sin 65° = \frac{\text{height}}{20}$ $\text{height} = 20 \sin 65° = 20(0.9063) \approx 18.13 \text{ ft}$

Step 3: Find the distance from wall (adjacent): $\cos 65° = \frac{\text{distance}}{20}$ $\text{distance} = 20 \cos 65° = 20(0.4226) \approx 8.45 \text{ ft}$

Check: $18.13^2 + 8.45^2 \approx 328.7 + 71.4 = 400.1 \approx 20^2$ ✓

Answer: Height ≈ 18.13 ft, Distance from wall ≈ 8.45 ft

Next Topic →

Complex Numbers on the SAT

🎴

Practice with Flashcards

Review key concepts with our flashcard system

📖

Browse All Topics

Explore more SAT Prep topics