Digital SAT (H-S)

🔺 Right Triangles and Trigonometry

1. Pythagorean Theorem

For any right triangle:

a2+b2=c2a^2 + b^2 = c^2a2+b2=c2

Where:

• $a$ and $b$ = legs of the triangle

• $c$ = hypotenuse (the side opposite the right angle)

Common Pythagorean Triples (helpful to memorize):

• $3,4,5$

• $5,12,13$

• $7,24,25$

• Multiples of these (e.g. $6,8,10$)

2. Trigonometric Ratios

For a right triangle with an acute angle $\theta$:

• sin⁡θ=oppositehypotenuse\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}sinθ=hypotenuseopposite​

• cos⁡θ=adjacenthypotenuse\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}cosθ=hypotenuseadjacent​

• tan⁡θ=oppositeadjacent\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}tanθ=adjacentopposite​

Mnemonic: SOH-CAH-TOA

3. Special Right Triangles

a) 45°-45°-90° Triangle

• Legs are equal

• Hypotenuse = leg × 2\sqrt{2}2​

b) 30°-60°-90° Triangle

• Short leg (opposite 30°): $x$

• Long leg (opposite 60°): x3x\sqrt{3}x3​

• Hypotenuse (opposite 90°): $2x$

4. Trigonometric Applications

• Used to find missing side lengths or angles

• Use inverse trig functions to find angles:

  • θ=sin⁡−1(opposite/hypotenuse)\theta = \sin^{-1}(\text{opposite}/\text{hypotenuse})θ=sin−1(opposite/hypotenuse)

  • θ=cos⁡−1(adjacent/hypotenuse)\theta = \cos^{-1}(\text{adjacent}/\text{hypotenuse})θ=cos−1(adjacent/hypotenuse)

  • θ=tan⁡−1(opposite/adjacent)\theta = \tan^{-1}(\text{opposite}/\text{adjacent})θ=tan−1(opposite/adjacent)

5. Word Problems

Look for:

• Right triangle setup (e.g. ladders, ramps, angles of elevation/depression)

• Use trig ratios or the Pythagorean Theorem