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)

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

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

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

Answer: The other leg is 8.

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

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

Answer: The other leg is 8.

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

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

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

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

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.

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.

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.

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

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

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