Right Triangles & Trigonometry – SAT Math

1. Pythagorean Theorem

\( a^2 + b^2 = c^2 \)

📌 RULE: Pythagorean Theorem

In a right triangle, the square of the hypotenuse (\(c\)) equals the sum of the squares of the other two sides (\(a\) and \(b\)).

\[ c^2 = a^2 + b^2 \]
where \(c\) is the hypotenuse (the side opposite the right angle).

Common Pythagorean Triples:
\(3-4-5\), \(5-12-13\), \(7-24-25\), \(8-15-17\), \(9-40-41\)

💡 Strategy — Pythagorean Theorem

1. Identify the hypotenuse (the side opposite the right angle).
2. Plug into \(a^2 + b^2 = c^2\).
3. Solve for the unknown side.

📝 SOLVED EXAMPLE 1 — Pythagorean Theorem

Find the hypotenuse of a right triangle with legs 6 and 8.

Step 1: \(a = 6\), \(b = 8\)

Step 2: \(c^2 = 6^2 + 8^2\)

Step 3: \(c^2 = 36 + 64 = 100\)

Step 4: \(c = \sqrt{100} = \color{var(--math)}{10}\)

✅ Hypotenuse = 10

💡 Tip: 6-8-10 is a multiple of the 3-4-5 triple.

2. Special Right Triangles

📌 RULE: 45°-45°-90° Triangle

Ratio of sides: \(1 : 1 : \sqrt{2}\)

\[ \text{Hypotenuse} = \text{Leg} \times \sqrt{2} \]
\[ \text{Leg} = \frac{\text{Hypotenuse}}{\sqrt{2}} \]

Example: Leg = 5 → Hypotenuse = \(5\sqrt{2}\)

📌 RULE: 30°-60°-90° Triangle

Ratio of sides: \(1 : \sqrt{3} : 2\)

\[ \text{Hypotenuse} = 2 \times \text{Short Leg} \]
\[ \text{Long Leg} = \text{Short Leg} \times \sqrt{3} \]

Example: Short leg = 4 → Long leg = \(4\sqrt{3}\), Hypotenuse = 8

📝 SOLVED EXAMPLE 2 — 45-45-90 Triangle

In a 45°-45°-90° triangle, the hypotenuse is \(10\sqrt{2}\). Find the length of each leg.

Step 1: Hypotenuse = Leg × \(\sqrt{2}\)

Step 2: \(10\sqrt{2} = L \times \sqrt{2}\)

Step 3: \(L = \color{var(--math)}{10}\)

✅ Each leg = 10

💡 Tip: In a 45-45-90 triangle, the legs are equal.

📝 SOLVED EXAMPLE 3 — 30-60-90 Triangle

In a 30°-60°-90° triangle, the short leg is 5. Find the long leg and the hypotenuse.

Step 1: Short leg = 5

Step 2: Long leg = 5 × \(\sqrt{3} = \color{var(--math)}{5\sqrt{3}}\)

Step 3: Hypotenuse = 2 × 5 = \color{var(--math)}{10}

✅ Long leg = \(5\sqrt{3}\), Hypotenuse = 10

💡 Tip: Hypotenuse is always twice the short leg.

3. SOH CAH TOA (Trigonometry)

Sine

\(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\)

SOH

Cosine

\(\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)

CAH

Tangent

\(\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}\)

TOA

📌 Key Concept

SOH CAH TOA is the mnemonic to remember trigonometric ratios:
Sine = Opposite / Hypotenuse
Cosine = Adjacent / Hypotenuse
Tangent = Opposite / Adjacent

📌 RULE: Using SOH CAH TOA

Step 1: Identify the angle \(\theta\) you're working with.
Step 2: Label the sides: Opposite (across from \(\theta\)), Adjacent (next to \(\theta\)), Hypotenuse (opposite the right angle).
Step 3: Choose the correct ratio based on the given information.
Step 4: Solve for the unknown.

📝 SOLVED EXAMPLE 4 — Sine

In a right triangle, the hypotenuse is 10 and \(\sin \theta = 0.6\). Find the opposite side.

Step 1: \(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\)

Step 2: \(0.6 = \frac{\text{Opposite}}{10}\)

Step 3: Opposite = \(0.6 \times 10 = \color{var(--math)}{6}\)

✅ Opposite = 6

💡 Tip: SOH → Sine = Opposite / Hypotenuse.

📝 SOLVED EXAMPLE 5 — Tangent

In a right triangle, the opposite side is 8 and the adjacent side is 6. Find \(\tan \theta\).

Step 1: \(\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}\)

Step 2: \(\tan \theta = \frac{8}{6} = \color{var(--math)}{\frac{4}{3}}\)

✅ \(\tan \theta = \frac{4}{3}\)

💡 Tip: TOA → Tangent = Opposite / Adjacent.

🧪 Practice Questions

Solve each problem using the rules above. Click "Show Answer" to see the full solution.

Question 1

Find the hypotenuse of a right triangle with legs 9 and 12.

A) 13

B) 14

C) 15

D) 16

✓ Answer: C

\(c^2 = 9^2 + 12^2 = 81 + 144 = 225\) → \(c = 15\)

📝 Solution: \(c = \sqrt{225} = 15\)

Question 2

In a 45°-45°-90° triangle, each leg is 7. What is the hypotenuse?

A) 7

B) 7√2

C) 14

D) 7√2

✓ Answer: D

Hypotenuse = Leg × √2 = 7√2

📝 Solution: \(7\sqrt{2}\)

Question 3

In a 30°-60°-90° triangle, the short leg is 6. What is the hypotenuse?

A) 6

B) 6√3

C) 12

D) 12√3

✓ Answer: C

Hypotenuse = 2 × short leg = 2 × 6 = 12

📝 Solution: \(2 \times 6 = 12\)

Question 4

In a right triangle, \(\sin \theta = \frac{3}{5}\). If the hypotenuse is 10, what is the opposite side?

A) 4

B) 6

C) 8

D) 9

✓ Answer: B

\(\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\) → \(\frac{3}{5} = \frac{x}{10}\) → \(x = 6\)

📝 Solution: \(3/5 = x/10\) → \(x = 6\)

Question 5

In a right triangle, \(\cos \theta = \frac{4}{5}\). If the hypotenuse is 15, what is the adjacent side?

A) 9

B) 10

C) 11

D) 12

✓ Answer: D

\(\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}\) → \(\frac{4}{5} = \frac{x}{15}\) → \(x = 12\)

📝 Solution: \(4/5 = x/15\) → \(x = 12\)

Question 6

Find the missing leg: a right triangle has hypotenuse 13 and one leg 5.

A) 10

B) 11

C) 12

D) 12

✓ Answer: D

\(13^2 = 5^2 + b^2\) → \(169 = 25 + b^2\) → \(b^2 = 144\) → \(b = 12\)

📝 Solution: 5-12-13 Pythagorean triple.

Question 7

In a 30°-60°-90° triangle, the long leg is \(9\sqrt{3}\). What is the short leg?

A) 3

B) 9

C) 12

D) 18

✓ Answer: B

Long leg = short leg × √3. So \(9\sqrt{3} = L\sqrt{3}\) → \(L = 9\)

📝 Solution: \(L = 9\)

Question 8

In a right triangle, \(\tan \theta = \frac{4}{3}\). What is \(\sin \theta\)?

A) \(\frac{3}{5}\)

B) \(\frac{4}{5}\)

C) \(\frac{4}{5}\)

D) \(\frac{5}{4}\)

✓ Answer: C

\(\tan \theta = \frac{4}{3}\) → opposite = 4, adjacent = 3. Hypotenuse = 5 (3-4-5 triple). \(\sin \theta = \frac{4}{5}\)

📝 Solution: 3-4-5 triangle → \(\sin \theta = \frac{4}{5}\)

Question 9

In a 45°-45°-90° triangle, the hypotenuse is \(14\sqrt{2}\). Find the length of each leg.

A) 7

B) 7√2

C) 14

D) 14

✓ Answer: D

Hypotenuse = Leg × √2 → \(14\sqrt{2} = L \times \sqrt{2}\) → \(L = 14\)

📝 Solution: \(L = 14\)

Question 10

In a right triangle, \(\cos \theta = \frac{12}{13}\). Find \(\sin \theta\).

A) \(\frac{5}{13}\)

B) \(\frac{5}{13}\)

C) \(\frac{12}{13}\)

D) \(\frac{13}{5}\)

✓ Answer: B

\(\cos \theta = \frac{12}{13}\) → adjacent = 12, hypotenuse = 13. Opposite = 5 (5-12-13 triple). \(\sin \theta = \frac{5}{13}\)

📝 Solution: 5-12-13 triangle → \(\sin \theta = \frac{5}{13}\)

🎉 WELL DONE!

You've completed the Right Triangles & Trigonometry lesson. You now know the Pythagorean theorem, SOH CAH TOA, and special right triangles (30-60-90, 45-45-90).

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