Unit 8 Right Triangles And Trigonometry Answer Key

Let's delve into the world of right triangles and trigonometry, focusing on the concepts typically covered in a "Unit 8" curriculum. This exploration will provide a solid understanding of the relationships between angles and sides in right triangles, and how trigonometric functions are used to solve for unknown values. While an "answer key" per se is not provided within this article, the comprehensive explanations and examples will equip you with the knowledge to solve related problems confidently.

Right Triangles: The Foundation

A right triangle is a triangle that contains one angle of exactly 90 degrees. This angle is often marked with a small square. The side opposite the right angle is called the hypotenuse, which is always the longest side of the triangle. The other two sides are called legs or cathetus. These are often referred to as the opposite and adjacent sides relative to a specific acute angle within the right triangle.

The Pythagorean Theorem: a² + b² = c²

One of the most fundamental theorems related to right triangles is the Pythagorean Theorem. It states that the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). This theorem is expressed as:

a² + b² = c²

This theorem allows you to find the length of any side of a right triangle if you know the lengths of the other two sides.

Example:

Suppose a right triangle has legs with lengths of 3 and 4. To find the length of the hypotenuse:

1. a = 3, b = 4, c = ?
2. 3² + 4² = c²
3. 9 + 16 = c²
4. 25 = c²
5. c = √25 = 5

Therefore, the length of the hypotenuse is 5.

Special Right Triangles

Certain right triangles appear frequently and have specific side ratios that are worth memorizing. These are often called "special right triangles".

• 45-45-90 Triangle: This triangle has angles of 45°, 45°, and 90°. The sides are in the ratio x : x : x√2, where x is the length of each leg and x√2 is the length of the hypotenuse.

• 30-60-90 Triangle: This triangle has angles of 30°, 60°, and 90°. The sides are in the ratio x : x√3 : 2x, where x is the length of the side opposite the 30° angle, x√3 is the length of the side opposite the 60° angle, and 2x is the length of the hypotenuse.

Knowing these ratios can significantly speed up problem-solving.

Example (45-45-90):

If a 45-45-90 triangle has a leg of length 7, then the other leg is also 7, and the hypotenuse is 7√2.

Example (30-60-90):

If a 30-60-90 triangle has a side opposite the 30° angle with a length of 5, then the side opposite the 60° angle is 5√3, and the hypotenuse is 10.

Introduction to Trigonometry

Trigonometry is the study of the relationships between the angles and sides of triangles. In the context of right triangles, we use three primary trigonometric functions: sine (sin), cosine (cos), and tangent (tan). These functions relate an acute angle in a right triangle to the ratio of two of its sides.

SOH CAH TOA

A helpful mnemonic for remembering these relationships is SOH CAH TOA:

• SOH: Sine = Opposite / Hypotenuse
• CAH: Cosine = Adjacent / Hypotenuse
• TOA: Tangent = Opposite / Adjacent

Definitions:

• Opposite: The side opposite to the angle you are considering.
• Adjacent: The side adjacent to the angle you are considering (not the hypotenuse).
• Hypotenuse: The side opposite the right angle (always the longest side).

Using Trigonometric Functions

To use these functions, you need to identify the angle you're working with, and then determine which sides are opposite, adjacent, and the hypotenuse relative to that angle.

Example:

Consider a right triangle with angle θ.

• sin(θ) = (Length of Opposite Side) / (Length of Hypotenuse)
• cos(θ) = (Length of Adjacent Side) / (Length of Hypotenuse)
• tan(θ) = (Length of Opposite Side) / (Length of Adjacent Side)

Example Problem:

In a right triangle, angle A is 30 degrees, and the hypotenuse is 10. Find the length of the side opposite angle A.

1. We are given the angle and the hypotenuse and want to find the opposite side. This means we should use the sine function (SOH).
2. sin(30°) = Opposite / 10
3. We know sin(30°) = 1/2
4. 1/2 = Opposite / 10
5. Opposite = 10 * (1/2) = 5

Therefore, the length of the side opposite angle A is 5.

Inverse Trigonometric Functions

Sometimes you know the ratio of the sides but need to find the angle. This is where inverse trigonometric functions come in. These are also known as arc functions.

• arcsin (or sin⁻¹): If sin(θ) = x, then arcsin(x) = θ
• arccos (or cos⁻¹): If cos(θ) = x, then arccos(x) = θ
• arctan (or tan⁻¹): If tan(θ) = x, then arctan(x) = θ

Example:

In a right triangle, the opposite side is 4, and the hypotenuse is 5. Find the angle θ.

1. sin(θ) = 4/5 = 0.8
2. θ = arcsin(0.8)
3. Using a calculator, θ ≈ 53.13°

Therefore, the angle θ is approximately 53.13 degrees.

Applications of Right Triangles and Trigonometry

Right triangles and trigonometry have countless applications in various fields. Here are a few examples:

• Navigation: Determining distances and directions using angles and trigonometric functions.
• Engineering: Designing structures, bridges, and buildings by calculating forces and angles.
• Physics: Analyzing projectile motion, wave phenomena, and optics.
• Surveying: Measuring land and creating maps.
• Computer Graphics: Creating realistic 3D models and animations.

Solving Real-World Problems

Many real-world problems can be modeled using right triangles. Here are a few examples:

• Angle of Elevation: The angle formed between the horizontal line of sight and an upward line of sight to an object.
• Angle of Depression: The angle formed between the horizontal line of sight and a downward line of sight to an object.

Example Problem (Angle of Elevation):

A building is 50 meters tall. From a point on the ground, the angle of elevation to the top of the building is 60 degrees. How far is the point on the ground from the base of the building?

1. We can draw a right triangle where the height of the building is the opposite side, the distance from the base of the building is the adjacent side, and the angle of elevation is 60 degrees.
2. tan(60°) = Opposite / Adjacent
3. tan(60°) = 50 / Adjacent
4. We know tan(60°) = √3
5. √3 = 50 / Adjacent
6. Adjacent = 50 / √3 ≈ 28.87 meters

Therefore, the point on the ground is approximately 28.87 meters from the base of the building.

Example Problem (Angle of Depression):

A lighthouse is 20 meters tall. A boat is observed from the top of the lighthouse with an angle of depression of 30 degrees. How far is the boat from the base of the lighthouse?

1. We can draw a right triangle where the height of the lighthouse is the opposite side (from the perspective of the boat's angle of elevation), the distance from the base of the lighthouse to the boat is the adjacent side, and the angle of depression is 30 degrees.
2. tan(30°) = Opposite / Adjacent
3. tan(30°) = 20 / Adjacent
4. We know tan(30°) = 1/√3
5. 1/√3 = 20 / Adjacent
6. Adjacent = 20 * √3 ≈ 34.64 meters

Therefore, the boat is approximately 34.64 meters from the base of the lighthouse.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variables for which the expressions are defined. These identities are crucial for simplifying trigonometric expressions and solving trigonometric equations. Here are some of the most common and useful identities:

Pythagorean Identities

These identities are derived from the Pythagorean Theorem:

• sin²(θ) + cos²(θ) = 1
• 1 + tan²(θ) = sec²(θ)
• 1 + cot²(θ) = csc²(θ)

Reciprocal Identities

These identities define the reciprocal trigonometric functions:

• csc(θ) = 1 / sin(θ)
• sec(θ) = 1 / cos(θ)
• cot(θ) = 1 / tan(θ)

Quotient Identities

These identities relate tangent and cotangent to sine and cosine:

• tan(θ) = sin(θ) / cos(θ)
• cot(θ) = cos(θ) / sin(θ)

Angle Sum and Difference Identities

These identities express trigonometric functions of sums and differences of angles:

• sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
• sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
• cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
• cos(A - B) = cos(A)cos(B) + sin(A)sin(B)
• tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A)tan(B))
• tan(A - B) = (tan(A) - tan(B)) / (1 + tan(A)tan(B))

Double-Angle Identities

These identities express trigonometric functions of double angles:

• sin(2θ) = 2sin(θ)cos(θ)
• cos(2θ) = cos²(θ) - sin²(θ) = 2cos²(θ) - 1 = 1 - 2sin²(θ)
• tan(2θ) = (2tan(θ)) / (1 - tan²(θ))

Half-Angle Identities

These identities express trigonometric functions of half angles:

• sin(θ/2) = ±√((1 - cos(θ)) / 2)
• cos(θ/2) = ±√((1 + cos(θ)) / 2)
• tan(θ/2) = ±√((1 - cos(θ)) / (1 + cos(θ))) = sin(θ) / (1 + cos(θ)) = (1 - cos(θ)) / sin(θ)

Using Trigonometric Identities to Simplify Expressions:

Trigonometric identities are often used to simplify complex trigonometric expressions.

Example:

Simplify the expression: (sin²(θ) + cos²(θ)) / cos(θ)

1. We know sin²(θ) + cos²(θ) = 1
2. Therefore, the expression simplifies to 1 / cos(θ)
3. We also know 1 / cos(θ) = sec(θ)
4. So, the simplified expression is sec(θ).

Solving Trigonometric Equations

Solving trigonometric equations involves finding the values of the variable (usually an angle) that make the equation true. This often involves using trigonometric identities to simplify the equation and then using inverse trigonometric functions to find the solutions.

General Approach:

1. Simplify the equation: Use trigonometric identities to simplify the equation as much as possible.
2. Isolate the trigonometric function: Get the trigonometric function (sin, cos, tan, etc.) by itself on one side of the equation.
3. Find the reference angle: Use inverse trigonometric functions to find the reference angle (the angle in the first quadrant that has the same trigonometric value).
4. Determine the quadrants: Determine the quadrants in which the solutions lie based on the sign of the trigonometric function. Remember "All Students Take Calculus" (ASTC):
   • All: All trigonometric functions are positive in Quadrant I.
   • Sine: Sine is positive in Quadrant II.
   • Tangent: Tangent is positive in Quadrant III.
   • Cosine: Cosine is positive in Quadrant IV.
5. Find all solutions within the desired interval: Use the reference angle and the quadrant information to find all solutions within the specified interval (usually 0 to 2π or 0 to 360 degrees). Remember to add multiples of 2π (or 360 degrees) to find all possible solutions.

Example:

Solve the equation sin(θ) = 1/2 for 0 ≤ θ < 2π.

1. The equation is already simplified and the trigonometric function is isolated.
2. Find the reference angle: θ_ref = arcsin(1/2) = π/6 (or 30 degrees).
3. Determine the quadrants: Sine is positive in Quadrants I and II.
4. Find the solutions:
   • Quadrant I: θ = θ_ref = π/6
   • Quadrant II: θ = π - θ_ref = π - π/6 = 5π/6
5. Therefore, the solutions are θ = π/6 and θ = 5π/6.

Example:

Solve the equation 2cos(θ) - 1 = 0 for 0 ≤ θ < 2π.

1. Simplify and isolate the trigonometric function:
   • 2cos(θ) = 1
   • cos(θ) = 1/2
2. Find the reference angle: θ_ref = arccos(1/2) = π/3 (or 60 degrees).
3. Determine the quadrants: Cosine is positive in Quadrants I and IV.
4. Find the solutions:
   • Quadrant I: θ = θ_ref = π/3
   • Quadrant IV: θ = 2π - θ_ref = 2π - π/3 = 5π/3
5. Therefore, the solutions are θ = π/3 and θ = 5π/3.

The Law of Sines and Law of Cosines

While the basic trigonometric functions (SOH CAH TOA) apply to right triangles, the Law of Sines and the Law of Cosines are used to solve any triangle (right, acute, or obtuse).

The Law of Sines

The Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is constant for all three sides and angles in a triangle.

• a / sin(A) = b / sin(B) = c / sin(C)

Where:

• a, b, and c are the lengths of the sides of the triangle.
• A, B, and C are the angles opposite those sides, respectively.

The Law of Sines is useful when you know:

• Two angles and one side (AAS or ASA).
• Two sides and an angle opposite one of those sides (SSA - this case can be ambiguous).

The Law of Cosines

The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles.

• a² = b² + c² - 2bc * cos(A)
• b² = a² + c² - 2ac * cos(B)
• c² = a² + b² - 2ab * cos(C)

The Law of Cosines is useful when you know:

• Three sides (SSS).
• Two sides and the included angle (SAS).

Example (Law of Sines):

In triangle ABC, A = 30°, B = 45°, and a = 10. Find the length of side b.

1. Using the Law of Sines: a / sin(A) = b / sin(B)
2. 10 / sin(30°) = b / sin(45°)
3. 10 / (1/2) = b / (√2/2)
4. 20 = b / (√2/2)
5. b = 20 * (√2/2) = 10√2

Therefore, the length of side b is 10√2.

Example (Law of Cosines):

In triangle ABC, a = 5, b = 7, and C = 60°. Find the length of side c.

1. Using the Law of Cosines: c² = a² + b² - 2ab * cos(C)
2. c² = 5² + 7² - 2 * 5 * 7 * cos(60°)
3. c² = 25 + 49 - 70 * (1/2)
4. c² = 74 - 35
5. c² = 39
6. c = √39

Therefore, the length of side c is √39.

Conclusion

This comprehensive overview has covered the essential concepts of right triangles and trigonometry, including the Pythagorean Theorem, special right triangles, trigonometric functions (sine, cosine, tangent), inverse trigonometric functions, trigonometric identities, solving trigonometric equations, and the Law of Sines and Law of Cosines. By understanding these principles and practicing problem-solving, you can develop a strong foundation in trigonometry and confidently tackle a wide range of mathematical challenges. Remember to utilize resources like textbooks, online tutorials, and practice problems to solidify your understanding and build your skills.