Right Triangle Trigonometry - Tenth Grade
Introduction to Trigonometry
Trigonometry: The study of relationships between angles and sides of triangles
Greek Origin: "trigonon" (triangle) + "metron" (measure)
Applies to: RIGHT triangles (triangles with one 90° angle)
Key Components:
• Hypotenuse: Side opposite the right angle (longest side)
• Opposite Side: Side opposite the angle of interest
• Adjacent Side: Side next to the angle of interest (not the hypotenuse)
Greek Origin: "trigonon" (triangle) + "metron" (measure)
Applies to: RIGHT triangles (triangles with one 90° angle)
Key Components:
• Hypotenuse: Side opposite the right angle (longest side)
• Opposite Side: Side opposite the angle of interest
• Adjacent Side: Side next to the angle of interest (not the hypotenuse)
1. Trigonometric Ratios: sin, cos, and tan
Trigonometric Ratios: Ratios of sides in a right triangle relative to an acute angle
Three Primary Ratios: Sine (sin), Cosine (cos), Tangent (tan)
Mnemonic: SOH-CAH-TOA
• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent
Three Primary Ratios: Sine (sin), Cosine (cos), Tangent (tan)
Mnemonic: SOH-CAH-TOA
• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent
Three Primary Trigonometric Ratios:
For angle θ in a right triangle:
1. SINE (sin):
$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{O}{H}$$
2. COSINE (cos):
$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{A}{H}$$
3. TANGENT (tan):
$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{O}{A}$$
Relationship:
$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
For angle θ in a right triangle:
1. SINE (sin):
$$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{O}{H}$$
2. COSINE (cos):
$$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{A}{H}$$
3. TANGENT (tan):
$$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{O}{A}$$
Relationship:
$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$
Example 1: Find all three ratios
Right triangle has legs 3 and 4, hypotenuse 5. Find sin θ, cos θ, and tan θ for the angle opposite the side of length 3.
Opposite = 3, Adjacent = 4, Hypotenuse = 5
$$\sin \theta = \frac{3}{5} = 0.6$$
$$\cos \theta = \frac{4}{5} = 0.8$$
$$\tan \theta = \frac{3}{4} = 0.75$$
Right triangle has legs 3 and 4, hypotenuse 5. Find sin θ, cos θ, and tan θ for the angle opposite the side of length 3.
Opposite = 3, Adjacent = 4, Hypotenuse = 5
$$\sin \theta = \frac{3}{5} = 0.6$$
$$\cos \theta = \frac{4}{5} = 0.8$$
$$\tan \theta = \frac{3}{4} = 0.75$$
2. Trigonometric Ratios with Radicals
Radicals in Trigonometry: When side lengths involve square roots (√)
Common in: Special right triangles and Pythagorean theorem problems
Key Skill: Simplifying radical expressions in ratios
Common in: Special right triangles and Pythagorean theorem problems
Key Skill: Simplifying radical expressions in ratios
Example 1: Triangle with radical sides
Right triangle has legs 1 and √3, hypotenuse 2. Find sin θ, cos θ, tan θ for angle opposite side 1.
Opposite = 1, Adjacent = √3, Hypotenuse = 2
$$\sin \theta = \frac{1}{2}$$
$$\cos \theta = \frac{\sqrt{3}}{2}$$
$$\tan \theta = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ (rationalized)
Note: To rationalize $\frac{1}{\sqrt{3}}$, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$
Right triangle has legs 1 and √3, hypotenuse 2. Find sin θ, cos θ, tan θ for angle opposite side 1.
Opposite = 1, Adjacent = √3, Hypotenuse = 2
$$\sin \theta = \frac{1}{2}$$
$$\cos \theta = \frac{\sqrt{3}}{2}$$
$$\tan \theta = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$ (rationalized)
Note: To rationalize $\frac{1}{\sqrt{3}}$, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$
Example 2: 45-45-90 triangle
In a 45-45-90 triangle with legs of length 1, find the trig ratios for 45°.
Legs = 1, Hypotenuse = √2
$$\sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\tan 45° = \frac{1}{1} = 1$$
In a 45-45-90 triangle with legs of length 1, find the trig ratios for 45°.
Legs = 1, Hypotenuse = √2
$$\sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$
$$\tan 45° = \frac{1}{1} = 1$$
3. Trigonometric Ratios: csc, sec, and cot
Reciprocal Ratios: The reciprocals of sin, cos, and tan
Three Reciprocal Ratios:
• Cosecant (csc): Reciprocal of sine
• Secant (sec): Reciprocal of cosine
• Cotangent (cot): Reciprocal of tangent
Three Reciprocal Ratios:
• Cosecant (csc): Reciprocal of sine
• Secant (sec): Reciprocal of cosine
• Cotangent (cot): Reciprocal of tangent
Three Reciprocal Trigonometric Ratios:
1. COSECANT (csc):
$$\csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{H}{O}$$
2. SECANT (sec):
$$\sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{H}{A}$$
3. COTANGENT (cot):
$$\cot \theta = \frac{1}{\tan \theta} = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{A}{O}$$
Relationships:
$$\cot \theta = \frac{\cos \theta}{\sin \theta}$$
1. COSECANT (csc):
$$\csc \theta = \frac{1}{\sin \theta} = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{H}{O}$$
2. SECANT (sec):
$$\sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{H}{A}$$
3. COTANGENT (cot):
$$\cot \theta = \frac{1}{\tan \theta} = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{A}{O}$$
Relationships:
$$\cot \theta = \frac{\cos \theta}{\sin \theta}$$
Example: Find reciprocal ratios
If sin θ = 3/5, cos θ = 4/5, and tan θ = 3/4, find csc θ, sec θ, and cot θ.
$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{3/5} = \frac{5}{3}$$
$$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{4/5} = \frac{5}{4}$$
$$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{3/4} = \frac{4}{3}$$
If sin θ = 3/5, cos θ = 4/5, and tan θ = 3/4, find csc θ, sec θ, and cot θ.
$$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{3/5} = \frac{5}{3}$$
$$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{4/5} = \frac{5}{4}$$
$$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{3/4} = \frac{4}{3}$$
4. Trigonometric Ratios in Similar Right Triangles
Key Concept: Trigonometric ratios are the SAME for a given angle, regardless of triangle size
Why: Similar triangles have proportional sides
Consequence: Trig ratios depend only on the angle, not the size of the triangle
Why: Similar triangles have proportional sides
Consequence: Trig ratios depend only on the angle, not the size of the triangle
Property of Similar Right Triangles:
If two right triangles are similar (same angles), then:
• All corresponding sides are proportional
• The trigonometric ratios for corresponding angles are EQUAL
• sin θ, cos θ, and tan θ are the same in both triangles
This is why we can create tables of trig values!
The value of sin 30° is always 0.5, no matter the triangle size.
If two right triangles are similar (same angles), then:
• All corresponding sides are proportional
• The trigonometric ratios for corresponding angles are EQUAL
• sin θ, cos θ, and tan θ are the same in both triangles
This is why we can create tables of trig values!
The value of sin 30° is always 0.5, no matter the triangle size.
Example: Similar triangles
Triangle 1: Sides 3-4-5, angle α opposite side 3
Triangle 2: Sides 6-8-10, angle β opposite side 6
Triangle 1:
$$\sin \alpha = \frac{3}{5} = 0.6$$
Triangle 2:
$$\sin \beta = \frac{6}{10} = 0.6$$
Same ratio! Because α = β (similar triangles)
Triangle 1: Sides 3-4-5, angle α opposite side 3
Triangle 2: Sides 6-8-10, angle β opposite side 6
Triangle 1:
$$\sin \alpha = \frac{3}{5} = 0.6$$
Triangle 2:
$$\sin \beta = \frac{6}{10} = 0.6$$
Same ratio! Because α = β (similar triangles)
5. Special Right Triangles
Special Right Triangles: Right triangles with specific angle measurements
Two Types:
• 45-45-90 Triangle: Isosceles right triangle
• 30-60-90 Triangle: Half of an equilateral triangle
Why Special: Side ratios are always the same, making calculations easy
Two Types:
• 45-45-90 Triangle: Isosceles right triangle
• 30-60-90 Triangle: Half of an equilateral triangle
Why Special: Side ratios are always the same, making calculations easy
45-45-90 Triangle
45-45-90 Triangle (Isosceles Right Triangle):
Angles: 45°, 45°, 90°
Side Ratio:
$$\text{Leg} : \text{Leg} : \text{Hypotenuse} = 1 : 1 : \sqrt{2}$$
If leg = x:
• Both legs = $x$
• Hypotenuse = $x\sqrt{2}$
If hypotenuse = h:
• Each leg = $\frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}$
Shortcut: $x : x : x\sqrt{2}$
Angles: 45°, 45°, 90°
Side Ratio:
$$\text{Leg} : \text{Leg} : \text{Hypotenuse} = 1 : 1 : \sqrt{2}$$
If leg = x:
• Both legs = $x$
• Hypotenuse = $x\sqrt{2}$
If hypotenuse = h:
• Each leg = $\frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}$
Shortcut: $x : x : x\sqrt{2}$
Example: 45-45-90 triangle
If leg = 5, find hypotenuse:
$$\text{Hypotenuse} = 5\sqrt{2} \approx 7.07$$
If hypotenuse = 10, find legs:
$$\text{Leg} = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07$$
If leg = 5, find hypotenuse:
$$\text{Hypotenuse} = 5\sqrt{2} \approx 7.07$$
If hypotenuse = 10, find legs:
$$\text{Leg} = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07$$
30-60-90 Triangle
30-60-90 Triangle (Half Equilateral Triangle):
Angles: 30°, 60°, 90°
Side Ratio:
$$\text{Short leg} : \text{Long leg} : \text{Hypotenuse} = 1 : \sqrt{3} : 2$$
If short leg (opposite 30°) = x:
• Short leg = $x$
• Long leg (opposite 60°) = $x\sqrt{3}$
• Hypotenuse = $2x$
Pattern: $x : x\sqrt{3} : 2x$
Remember:
• Short leg is opposite 30° angle
• Long leg is opposite 60° angle
• Hypotenuse is opposite 90° angle
Angles: 30°, 60°, 90°
Side Ratio:
$$\text{Short leg} : \text{Long leg} : \text{Hypotenuse} = 1 : \sqrt{3} : 2$$
If short leg (opposite 30°) = x:
• Short leg = $x$
• Long leg (opposite 60°) = $x\sqrt{3}$
• Hypotenuse = $2x$
Pattern: $x : x\sqrt{3} : 2x$
Remember:
• Short leg is opposite 30° angle
• Long leg is opposite 60° angle
• Hypotenuse is opposite 90° angle
Example: 30-60-90 triangle
If short leg = 4, find other sides:
• Long leg = $4\sqrt{3} \approx 6.93$
• Hypotenuse = $2(4) = 8$
If hypotenuse = 12, find legs:
• Short leg = $\frac{12}{2} = 6$
• Long leg = $6\sqrt{3} \approx 10.39$
If short leg = 4, find other sides:
• Long leg = $4\sqrt{3} \approx 6.93$
• Hypotenuse = $2(4) = 8$
If hypotenuse = 12, find legs:
• Short leg = $\frac{12}{2} = 6$
• Long leg = $6\sqrt{3} \approx 10.39$
6. Find Trigonometric Functions of Special Angles
Special Angles: 0°, 30°, 45°, 60°, 90°
Why Important: Exact values (no calculator needed)
Based on: Special right triangles and unit circle
Why Important: Exact values (no calculator needed)
Based on: Special right triangles and unit circle
Trig Values for Special Angles (MEMORIZE!):
| Angle θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ |
| 45° | $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ | $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ | 1 |
| 60° | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
| 90° | 1 | 0 | undefined |
Memory Tricks:
For Sine (0° to 90°):
$$\sin 0° = \sqrt{\frac{0}{4}} = 0$$
$$\sin 30° = \sqrt{\frac{1}{4}} = \frac{1}{2}$$
$$\sin 45° = \sqrt{\frac{2}{4}} = \frac{\sqrt{2}}{2}$$
$$\sin 60° = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$
$$\sin 90° = \sqrt{\frac{4}{4}} = 1$$
For Cosine: Reverse the sine values!
cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = 1/2, cos 90° = 0
For Sine (0° to 90°):
$$\sin 0° = \sqrt{\frac{0}{4}} = 0$$
$$\sin 30° = \sqrt{\frac{1}{4}} = \frac{1}{2}$$
$$\sin 45° = \sqrt{\frac{2}{4}} = \frac{\sqrt{2}}{2}$$
$$\sin 60° = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$
$$\sin 90° = \sqrt{\frac{4}{4}} = 1$$
For Cosine: Reverse the sine values!
cos 0° = 1, cos 30° = √3/2, cos 45° = √2/2, cos 60° = 1/2, cos 90° = 0
7. Find Trigonometric Functions Using a Calculator
Calculator Instructions:
Step 1: Check Mode
• Make sure calculator is in DEGREE mode (not radians)
• Look for "DEG" or "D" on screen
Step 2: Enter Function
• Press sin, cos, or tan button
• Enter angle in degrees
• Press enter or equals
Examples:
• sin(25°) ≈ 0.4226
• cos(70°) ≈ 0.3420
• tan(50°) ≈ 1.1918
Step 1: Check Mode
• Make sure calculator is in DEGREE mode (not radians)
• Look for "DEG" or "D" on screen
Step 2: Enter Function
• Press sin, cos, or tan button
• Enter angle in degrees
• Press enter or equals
Examples:
• sin(25°) ≈ 0.4226
• cos(70°) ≈ 0.3420
• tan(50°) ≈ 1.1918
Important Calculator Notes:
• Degrees vs. Radians: Always check mode!
• Rounding: Use at least 4 decimal places for accuracy
• Special Angles: Calculator gives decimals; exact values are preferred
• Error Message: Tan 90° gives error (undefined)
• Verify: Check answers make sense (sin and cos are between -1 and 1)
• Degrees vs. Radians: Always check mode!
• Rounding: Use at least 4 decimal places for accuracy
• Special Angles: Calculator gives decimals; exact values are preferred
• Error Message: Tan 90° gives error (undefined)
• Verify: Check answers make sense (sin and cos are between -1 and 1)
8. Sine and Cosine of Complementary Angles
Complementary Angles: Two angles that add to 90°
In Right Triangle: The two acute angles are always complementary
Key Relationship: sin(θ) = cos(90° - θ)
In Right Triangle: The two acute angles are always complementary
Key Relationship: sin(θ) = cos(90° - θ)
Cofunction Identities (Complementary Angles):
If angles A and B are complementary (A + B = 90°), then:
$$\sin A = \cos B$$
$$\cos A = \sin B$$
$$\tan A = \cot B$$
Equivalently:
$$\sin \theta = \cos(90° - \theta)$$
$$\cos \theta = \sin(90° - \theta)$$
$$\tan \theta = \cot(90° - \theta)$$
Also:
$$\csc \theta = \sec(90° - \theta)$$
$$\sec \theta = \csc(90° - \theta)$$
$$\cot \theta = \tan(90° - \theta)$$
If angles A and B are complementary (A + B = 90°), then:
$$\sin A = \cos B$$
$$\cos A = \sin B$$
$$\tan A = \cot B$$
Equivalently:
$$\sin \theta = \cos(90° - \theta)$$
$$\cos \theta = \sin(90° - \theta)$$
$$\tan \theta = \cot(90° - \theta)$$
Also:
$$\csc \theta = \sec(90° - \theta)$$
$$\sec \theta = \csc(90° - \theta)$$
$$\cot \theta = \tan(90° - \theta)$$
Example 1:
$$\sin 30° = \cos 60°$$
Both equal $\frac{1}{2}$ ✓
$$\sin 45° = \cos 45°$$
Both equal $\frac{\sqrt{2}}{2}$ ✓
$$\sin 30° = \cos 60°$$
Both equal $\frac{1}{2}$ ✓
$$\sin 45° = \cos 45°$$
Both equal $\frac{\sqrt{2}}{2}$ ✓
Example 2:
If sin 25° = 0.4226, find cos 65°
Since 25° + 65° = 90°:
$$\cos 65° = \sin 25° = 0.4226$$
If sin 25° = 0.4226, find cos 65°
Since 25° + 65° = 90°:
$$\cos 65° = \sin 25° = 0.4226$$
9. Inverse Trigonometric Functions
Inverse Trig Functions: Find the angle when given the ratio
Notation: sin⁻¹, cos⁻¹, tan⁻¹ OR arcsin, arccos, arctan
Read as: "inverse sine" or "arcsine"
Purpose: Reverse operation - from ratio to angle
Notation: sin⁻¹, cos⁻¹, tan⁻¹ OR arcsin, arccos, arctan
Read as: "inverse sine" or "arcsine"
Purpose: Reverse operation - from ratio to angle
Inverse Trigonometric Functions:
1. Inverse Sine (Arcsine):
$$\theta = \sin^{-1}(x) \text{ or } \theta = \arcsin(x)$$
If $\sin \theta = x$, then $\theta = \sin^{-1}(x)$
2. Inverse Cosine (Arccosine):
$$\theta = \cos^{-1}(x) \text{ or } \theta = \arccos(x)$$
If $\cos \theta = x$, then $\theta = \cos^{-1}(x)$
3. Inverse Tangent (Arctangent):
$$\theta = \tan^{-1}(x) \text{ or } \theta = \arctan(x)$$
If $\tan \theta = x$, then $\theta = \tan^{-1}(x)$
Domain Restrictions:
• sin⁻¹ and cos⁻¹: x must be between -1 and 1
• tan⁻¹: x can be any real number
1. Inverse Sine (Arcsine):
$$\theta = \sin^{-1}(x) \text{ or } \theta = \arcsin(x)$$
If $\sin \theta = x$, then $\theta = \sin^{-1}(x)$
2. Inverse Cosine (Arccosine):
$$\theta = \cos^{-1}(x) \text{ or } \theta = \arccos(x)$$
If $\cos \theta = x$, then $\theta = \cos^{-1}(x)$
3. Inverse Tangent (Arctangent):
$$\theta = \tan^{-1}(x) \text{ or } \theta = \arctan(x)$$
If $\tan \theta = x$, then $\theta = \tan^{-1}(x)$
Domain Restrictions:
• sin⁻¹ and cos⁻¹: x must be between -1 and 1
• tan⁻¹: x can be any real number
Example 1: Find angles
a) sin θ = 0.5, find θ
$$\theta = \sin^{-1}(0.5) = 30°$$
b) cos θ = 0.5, find θ
$$\theta = \cos^{-1}(0.5) = 60°$$
c) tan θ = 1, find θ
$$\theta = \tan^{-1}(1) = 45°$$
a) sin θ = 0.5, find θ
$$\theta = \sin^{-1}(0.5) = 30°$$
b) cos θ = 0.5, find θ
$$\theta = \cos^{-1}(0.5) = 60°$$
c) tan θ = 1, find θ
$$\theta = \tan^{-1}(1) = 45°$$
Example 2: Using calculator
Find θ if sin θ = 0.7
On calculator: sin⁻¹(0.7) ≈ 44.43°
Find θ if tan θ = 2.5
On calculator: tan⁻¹(2.5) ≈ 68.20°
Find θ if sin θ = 0.7
On calculator: sin⁻¹(0.7) ≈ 44.43°
Find θ if tan θ = 2.5
On calculator: tan⁻¹(2.5) ≈ 68.20°
10. Trigonometric Ratios: Find a Side Length
Steps to Find a Side Length:
Step 1: Identify the given angle and given side
Step 2: Identify the side you need to find
Step 3: Choose the right trig ratio:
• Use sin if you have/need opposite or hypotenuse
• Use cos if you have/need adjacent or hypotenuse
• Use tan if you have/need opposite or adjacent
Step 4: Set up equation
Step 5: Solve for unknown side
Step 1: Identify the given angle and given side
Step 2: Identify the side you need to find
Step 3: Choose the right trig ratio:
• Use sin if you have/need opposite or hypotenuse
• Use cos if you have/need adjacent or hypotenuse
• Use tan if you have/need opposite or adjacent
Step 4: Set up equation
Step 5: Solve for unknown side
Example 1: Find hypotenuse
Given: angle = 30°, opposite side = 5. Find hypotenuse.
Use sine (have opposite, need hypotenuse):
$$\sin 30° = \frac{5}{h}$$
$$0.5 = \frac{5}{h}$$
$$h = \frac{5}{0.5} = 10$$
Answer: Hypotenuse = 10
Given: angle = 30°, opposite side = 5. Find hypotenuse.
Use sine (have opposite, need hypotenuse):
$$\sin 30° = \frac{5}{h}$$
$$0.5 = \frac{5}{h}$$
$$h = \frac{5}{0.5} = 10$$
Answer: Hypotenuse = 10
Example 2: Find opposite side
Given: angle = 40°, adjacent = 12. Find opposite.
Use tangent (have adjacent, need opposite):
$$\tan 40° = \frac{x}{12}$$
$$0.8391 = \frac{x}{12}$$
$$x = 12 \times 0.8391 = 10.07$$
Answer: Opposite ≈ 10.07
Given: angle = 40°, adjacent = 12. Find opposite.
Use tangent (have adjacent, need opposite):
$$\tan 40° = \frac{x}{12}$$
$$0.8391 = \frac{x}{12}$$
$$x = 12 \times 0.8391 = 10.07$$
Answer: Opposite ≈ 10.07
11. Trigonometric Ratios: Find an Angle Measure
Steps to Find an Angle:
Step 1: Identify which two sides are given
Step 2: Choose the appropriate trig ratio:
• sin if you have opposite and hypotenuse
• cos if you have adjacent and hypotenuse
• tan if you have opposite and adjacent
Step 3: Calculate the ratio
Step 4: Use inverse function to find angle
Step 1: Identify which two sides are given
Step 2: Choose the appropriate trig ratio:
• sin if you have opposite and hypotenuse
• cos if you have adjacent and hypotenuse
• tan if you have opposite and adjacent
Step 3: Calculate the ratio
Step 4: Use inverse function to find angle
Example 1: Find angle using tan
Opposite = 7, Adjacent = 10. Find angle θ.
$$\tan \theta = \frac{7}{10} = 0.7$$
$$\theta = \tan^{-1}(0.7) \approx 35.0°$$
Opposite = 7, Adjacent = 10. Find angle θ.
$$\tan \theta = \frac{7}{10} = 0.7$$
$$\theta = \tan^{-1}(0.7) \approx 35.0°$$
Example 2: Find angle using sin
Opposite = 8, Hypotenuse = 15. Find angle θ.
$$\sin \theta = \frac{8}{15} = 0.5333$$
$$\theta = \sin^{-1}(0.5333) \approx 32.2°$$
Opposite = 8, Hypotenuse = 15. Find angle θ.
$$\sin \theta = \frac{8}{15} = 0.5333$$
$$\theta = \sin^{-1}(0.5333) \approx 32.2°$$
12. Solve a Right Triangle
Solving a Right Triangle: Finding ALL unknown sides and angles
Given Information: Usually one side and one angle (plus the right angle)
Need to Find: Two sides and one angle
Tools: Trig ratios, Pythagorean theorem, angle sum (180°)
Given Information: Usually one side and one angle (plus the right angle)
Need to Find: Two sides and one angle
Tools: Trig ratios, Pythagorean theorem, angle sum (180°)
Steps to Solve a Right Triangle:
Step 1: Draw and label the triangle
Step 2: Find the third angle
Use: Sum of angles in triangle = 180°
Third angle = 90° - given acute angle
Step 3: Find the second side
Use appropriate trig ratio
Step 4: Find the third side
Use another trig ratio OR Pythagorean theorem
Step 5: Check your work
Verify using Pythagorean theorem: $a^2 + b^2 = c^2$
Step 1: Draw and label the triangle
Step 2: Find the third angle
Use: Sum of angles in triangle = 180°
Third angle = 90° - given acute angle
Step 3: Find the second side
Use appropriate trig ratio
Step 4: Find the third side
Use another trig ratio OR Pythagorean theorem
Step 5: Check your work
Verify using Pythagorean theorem: $a^2 + b^2 = c^2$
Example: Complete solution
Given: Right triangle with ∠A = 90°, ∠B = 35°, side a (opposite A) = 20
Find: ∠C, side b, side c
Step 1: Find ∠C
$$\angle C = 180° - 90° - 35° = 55°$$
Step 2: Find side b (adjacent to ∠B)
$$\tan 35° = \frac{20}{b}$$
$$b = \frac{20}{\tan 35°} = \frac{20}{0.7002} \approx 28.57$$
Step 3: Find side c (hypotenuse)
$$\sin 35° = \frac{20}{c}$$
$$c = \frac{20}{\sin 35°} = \frac{20}{0.5736} \approx 34.87$$
Step 4: Verify
$$20^2 + 28.57^2 = 400 + 816.2 = 1216.2$$
$$34.87^2 = 1216.0$$ ✓ (close enough with rounding)
Answer: ∠C = 55°, b ≈ 28.57, c ≈ 34.87
Given: Right triangle with ∠A = 90°, ∠B = 35°, side a (opposite A) = 20
Find: ∠C, side b, side c
Step 1: Find ∠C
$$\angle C = 180° - 90° - 35° = 55°$$
Step 2: Find side b (adjacent to ∠B)
$$\tan 35° = \frac{20}{b}$$
$$b = \frac{20}{\tan 35°} = \frac{20}{0.7002} \approx 28.57$$
Step 3: Find side c (hypotenuse)
$$\sin 35° = \frac{20}{c}$$
$$c = \frac{20}{\sin 35°} = \frac{20}{0.5736} \approx 34.87$$
Step 4: Verify
$$20^2 + 28.57^2 = 400 + 816.2 = 1216.2$$
$$34.87^2 = 1216.0$$ ✓ (close enough with rounding)
Answer: ∠C = 55°, b ≈ 28.57, c ≈ 34.87
Trigonometric Ratios Summary
| Ratio | Formula | Reciprocal | Mnemonic |
|---|---|---|---|
| sin θ | $\frac{\text{Opposite}}{\text{Hypotenuse}}$ | csc θ | SOH |
| cos θ | $\frac{\text{Adjacent}}{\text{Hypotenuse}}$ | sec θ | CAH |
| tan θ | $\frac{\text{Opposite}}{\text{Adjacent}}$ | cot θ | TOA |
| csc θ | $\frac{\text{Hypotenuse}}{\text{Opposite}}$ | sin θ | - |
| sec θ | $\frac{\text{Hypotenuse}}{\text{Adjacent}}$ | cos θ | - |
| cot θ | $\frac{\text{Adjacent}}{\text{Opposite}}$ | tan θ | - |
Special Angles Reference Table
| Angle | sin | cos | tan | csc | sec | cot |
|---|---|---|---|---|---|---|
| 0° | 0 | 1 | 0 | undefined | 1 | undefined |
| 30° | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ | 2 | $\frac{2\sqrt{3}}{3}$ | $\sqrt{3}$ |
| 45° | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | 1 | $\sqrt{2}$ | $\sqrt{2}$ | 1 |
| 60° | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ | $\frac{2\sqrt{3}}{3}$ | 2 | $\frac{\sqrt{3}}{3}$ |
| 90° | 1 | 0 | undefined | 1 | undefined | 0 |
Special Right Triangles Summary
| Triangle Type | Angles | Side Ratio | If one side = x |
|---|---|---|---|
| 45-45-90 | 45°-45°-90° | $1 : 1 : \sqrt{2}$ | $x : x : x\sqrt{2}$ |
| 30-60-90 | 30°-60°-90° | $1 : \sqrt{3} : 2$ | $x : x\sqrt{3} : 2x$ |
Key Identities and Relationships
| Identity | Formula |
|---|---|
| Quotient Identity | $\tan \theta = \frac{\sin \theta}{\cos \theta}$ |
| Pythagorean Identity | $\sin^2 \theta + \cos^2 \theta = 1$ |
| Reciprocal Identities | $\csc \theta = \frac{1}{\sin \theta}$, $\sec \theta = \frac{1}{\cos \theta}$, $\cot \theta = \frac{1}{\tan \theta}$ |
| Cofunction Identities | $\sin \theta = \cos(90° - \theta)$, $\tan \theta = \cot(90° - \theta)$ |
Problem-Solving Strategy
| Goal | Given | Method |
|---|---|---|
| Find side | Angle + one side | Use sin, cos, or tan |
| Find angle | Two sides | Use sin⁻¹, cos⁻¹, or tan⁻¹ |
| Find hypotenuse | Two legs | Pythagorean theorem: $c = \sqrt{a^2 + b^2}$ |
| Solve triangle | One angle + one side | 1) Find third angle 2) Use trig to find sides 3) Verify with Pythagorean |
Success Tips for Right Triangle Trigonometry:
✓ SOH-CAH-TOA: sin = O/H, cos = A/H, tan = O/A
✓ MEMORIZE special angles: 0°, 30°, 45°, 60°, 90°
✓ 45-45-90 triangle: sides are $x : x : x\sqrt{2}$
✓ 30-60-90 triangle: sides are $x : x\sqrt{3} : 2x$
✓ Inverse functions find angles: use sin⁻¹, cos⁻¹, tan⁻¹
✓ Complementary angles: sin θ = cos(90° - θ)
✓ Always check calculator is in DEGREE mode!
✓ Reciprocals: csc = 1/sin, sec = 1/cos, cot = 1/tan
✓ To solve triangle: find third angle, then use trig for sides
✓ Verify answers with Pythagorean theorem: $a^2 + b^2 = c^2$
✓ SOH-CAH-TOA: sin = O/H, cos = A/H, tan = O/A
✓ MEMORIZE special angles: 0°, 30°, 45°, 60°, 90°
✓ 45-45-90 triangle: sides are $x : x : x\sqrt{2}$
✓ 30-60-90 triangle: sides are $x : x\sqrt{3} : 2x$
✓ Inverse functions find angles: use sin⁻¹, cos⁻¹, tan⁻¹
✓ Complementary angles: sin θ = cos(90° - θ)
✓ Always check calculator is in DEGREE mode!
✓ Reciprocals: csc = 1/sin, sec = 1/cos, cot = 1/tan
✓ To solve triangle: find third angle, then use trig for sides
✓ Verify answers with Pythagorean theorem: $a^2 + b^2 = c^2$