Solving Triangles Using Law Of Sines

Solving triangles, whether on paper or in practical applications like surveying or navigation, often requires us to determine unknown angles or side lengths. The Law of Sines is a powerful tool that enables us to tackle such problems, offering a straightforward relationship between the sides and angles of any triangle, not just right triangles. This article will delve deep into understanding and applying the Law of Sines, providing you with the knowledge and confidence to solve a wide range of triangle-related problems.

Understanding the Law of Sines

The Law of Sines states that for any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. Mathematically, this can be expressed as:

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 to sides a, b, and c, respectively.

This deceptively simple formula unlocks a world of possibilities when dealing with triangles. It allows us to determine unknown sides or angles if we know certain other values. The key lies in understanding when and how to apply this law effectively.

When to Use the Law of Sines

The Law of Sines is particularly useful when you are given one of the following scenarios:

• Angle-Side-Angle (ASA): You know two angles and the included side (the side between the two angles).
• Angle-Angle-Side (AAS): You know two angles and a non-included side (a side not between the two angles).
• Side-Side-Angle (SSA): You know two sides and an angle opposite one of those sides. This case is also known as the ambiguous case and requires extra care, as it might lead to zero, one, or two possible solutions.

Let's explore each of these cases in detail with illustrative examples.

Case 1: Angle-Side-Angle (ASA)

In the ASA case, we are given two angles and the side between them. The goal is to find the remaining angle and the other two sides.

Example:

Consider a triangle ABC where:

• Angle A = 30°
• Angle C = 70°
• Side b = 10

Steps to solve:

1. Find the missing angle: Since the sum of angles in a triangle is always 180°, we can find angle B:

   B = 180° - A - C = 180° - 30° - 70° = 80°

2. Apply the Law of Sines to find side a:

   a / sin(A) = b / sin(B)

   a / sin(30°) = 10 / sin(80°)

   a = (10 * sin(30°)) / sin(80°)

   a ≈ (10 * 0.5) / 0.9848

   a ≈ 5.077

3. Apply the Law of Sines to find side c:

   c / sin(C) = b / sin(B)

   c / sin(70°) = 10 / sin(80°)

   c = (10 * sin(70°)) / sin(80°)

   c ≈ (10 * 0.9397) / 0.9848

   c ≈ 9.542

Therefore, in triangle ABC, angle B ≈ 80°, side a ≈ 5.077, and side c ≈ 9.542.

Case 2: Angle-Angle-Side (AAS)

In the AAS case, we are given two angles and a side that is not between them. The process is very similar to the ASA case.

Example:

Consider a triangle ABC where:

• Angle A = 40°
• Angle B = 60°
• Side a = 8

Steps to solve:

1. Find the missing angle:

   C = 180° - A - B = 180° - 40° - 60° = 80°

2. Apply the Law of Sines to find side b:

   b / sin(B) = a / sin(A)

   b / sin(60°) = 8 / sin(40°)

   b = (8 * sin(60°)) / sin(40°)

   b ≈ (8 * 0.8660) / 0.6428

   b ≈ 10.77

3. Apply the Law of Sines to find side c:

   c / sin(C) = a / sin(A)

   c / sin(80°) = 8 / sin(40°)

   c = (8 * sin(80°)) / sin(40°)

   c ≈ (8 * 0.9848) / 0.6428

   c ≈ 12.25

Therefore, in triangle ABC, angle C ≈ 80°, side b ≈ 10.77, and side c ≈ 12.25.

Case 3: Side-Side-Angle (SSA) - The Ambiguous Case

The SSA case, also known as the ambiguous case, presents a unique challenge because the given information might lead to zero, one, or two possible triangles. This ambiguity arises because the given side opposite the angle can "swing" to create different triangles.

To determine the number of possible triangles, we need to analyze the relationship between the given sides and the angle. Let's denote the given angle as A, the side opposite to it as a, and the other given side as b.

Sub-Cases of SSA:

• Case 1: a < b * sin(A): No triangle exists. The side a is too short to reach the base.
• Case 2: a = b * sin(A): One right triangle exists. The side a is exactly the right length to form a right angle.
• Case 3: a > b: One triangle exists. The side a is long enough to ensure only one possible triangle can be formed.
• Case 4: b * sin(A) < a < b: Two triangles exist. This is the truly ambiguous case where two different triangles can be formed with the given information.

Example demonstrating the ambiguous case (two triangles):

Consider a triangle ABC where:

• Angle A = 30°
• Side a = 6
• Side b = 10

Steps to solve:

1. Check for the number of possible triangles:

   First, calculate b * sin(A):

   10 * sin(30°) = 10 * 0.5 = 5

   Since 5 < 6 < 10 (b * sin(A) < a < b), we have the ambiguous case, and two triangles are possible.

2. Apply the Law of Sines to find angle B:

   sin(B) / b = sin(A) / a

   sin(B) / 10 = sin(30°) / 6

   sin(B) = (10 * sin(30°)) / 6

   sin(B) = (10 * 0.5) / 6

   sin(B) ≈ 0.8333

   B1 = arcsin(0.8333) ≈ 56.44°

3. Find the second possible angle B (B2):

   Since sin(x) = sin(180° - x), we can find the second possible angle B:

   B2 = 180° - B1 ≈ 180° - 56.44° ≈ 123.56°

4. Solve for the remaining angles and sides for each triangle:

   • Triangle 1 (using B1 ≈ 56.44°):

     C1 = 180° - A - B1 ≈ 180° - 30° - 56.44° ≈ 93.56°

     c1 / sin(C1) = a / sin(A)

     c1 / sin(93.56°) = 6 / sin(30°)

     c1 = (6 * sin(93.56°)) / sin(30°)

     c1 ≈ (6 * 0.9981) / 0.5

     c1 ≈ 11.98

   • Triangle 2 (using B2 ≈ 123.56°):

     C2 = 180° - A - B2 ≈ 180° - 30° - 123.56° ≈ 26.44°

     c2 / sin(C2) = a / sin(A)

     c2 / sin(26.44°) = 6 / sin(30°)

     c2 = (6 * sin(26.44°)) / sin(30°)

     c2 ≈ (6 * 0.4456) / 0.5

     c2 ≈ 5.35

Therefore, in this SSA case, we have two possible triangles:

• Triangle 1: B1 ≈ 56.44°, C1 ≈ 93.56°, c1 ≈ 11.98
• Triangle 2: B2 ≈ 123.56°, C2 ≈ 26.44°, c2 ≈ 5.35

This example demonstrates the importance of carefully analyzing the SSA case to identify all possible solutions. Remember to always check if the given information leads to zero, one, or two triangles.

Practical Applications of the Law of Sines

The Law of Sines is not just a theoretical concept; it has numerous practical applications in various fields, including:

• Surveying: Determining distances and elevations in land surveying, especially in areas with irregular terrain.
• Navigation: Calculating distances and bearings in air and sea navigation. Pilots and sailors use the Law of Sines to determine their position and course.
• Astronomy: Measuring distances to stars and other celestial objects.
• Engineering: Analyzing forces in structures and designing bridges and other constructions.

Example: Navigation

A ship is sailing from point A to point B, which is 50 nautical miles away. The ship then turns 30 degrees towards point C. If the distance from B to C is 35 nautical miles, what is the distance from A to C?

1. Visualize the problem: Draw a triangle ABC where AB = 50, BC = 35, and angle ABC = 180° - 30° = 150°.

2. Apply the Law of Sines to find angle A:

   sin(A) / BC = sin(B) / AC (We'll call the distance from A to C 'x')

   sin(A) / 35 = sin(150°) / x (We don't know 'x' yet, so we'll find angle A first)

   To find angle A, we'll use: sin(A) / 35 = sin(150°) / x. But we need to find angle C first, since we can't directly solve for A without knowing AC (x). We'll rearrange the Law of Sines to help us find angle C:

   sin(C) / AB = sin(B) / AC -> sin(C) / 50 = sin(150°) / x. Still, we need to eliminate 'x'. This requires a different approach! We can use the Law of Cosines, which is better suited when we know two sides and the included angle.

   Let's switch to the Law of Cosines for a moment:

   AC² = AB² + BC² - 2(AB)(BC)cos(B)

   x² = 50² + 35² - 2(50)(35)cos(150°)

   x² = 2500 + 1225 - 3500 * (-0.8660)

   x² = 3725 + 3031

   x² ≈ 6756

   x ≈ √6756 ≈ 82.19 nautical miles.

3. Now we know AC (x) ≈ 82.19. Now we can find angle A (using Law of Sines):

sin(A) / 35 = sin(150°) / 82.19

sin(A) = (35 * sin(150°)) / 82.19

sin(A) = (35 * 0.5) / 82.19

sin(A) ≈ 0.213

A ≈ arcsin(0.213) ≈ 12.3°

4. Solve for the remaining angle

C = 180 - 150 - 12.3 = 17.7°

Therefore, the distance from A to C is approximately 82.19 nautical miles. The angle at A is approximately 12.3 degrees, and the angle at C is approximately 17.7 degrees. This example demonstrates how the Law of Sines (and sometimes, the Law of Cosines) can be used in real-world navigation problems.

Common Mistakes and How to Avoid Them

While the Law of Sines is a powerful tool, it's easy to make mistakes if you're not careful. Here are some common pitfalls and how to avoid them:

• Incorrectly identifying the SSA case: Always check the conditions for the ambiguous case before assuming a unique solution. Draw a diagram to visualize the possible triangles.
• Forgetting the second possible angle in the SSA case: When you find an angle using the inverse sine function (arcsin), remember that there might be a second possible angle in the range of 0° to 180°. Always check if both angles lead to valid triangles (i.e., the sum of the angles is less than 180°).
• Using the Law of Sines when the Law of Cosines is more appropriate: If you are given Side-Angle-Side (SAS) or Side-Side-Side (SSS), the Law of Cosines is the preferred method.
• Rounding errors: Avoid rounding intermediate calculations to maintain accuracy in your final answer. Keep as many decimal places as possible until the final step.
• Confusing angles and sides: Make sure you correctly identify which angle is opposite which side. Label your triangle clearly to avoid confusion.

Law of Sines: Proof

The Law of Sines can be proved using basic trigonometry and the area of a triangle.

Proof:

Consider any triangle ABC. Let's draw an altitude (height) from vertex B to side AC, and call the point where it intersects side AC as D. Let's denote the length of this altitude as h.

Now, we have two right triangles: ABD and CBD.

In right triangle ABD:

sin(A) = h / c => h = c * sin(A)

In right triangle CBD:

sin(C) = h / a => h = a * sin(C)

Since both expressions equal h, we can equate them:

c * sin(A) = a * sin(C)

Dividing both sides by sin(A) * sin(C), we get:

c / sin(C) = a / sin(A)

Similarly, we can draw an altitude from vertex A to side BC, and let's call the point where it intersects side BC as E. Let's denote the length of this altitude as h'.

In the right triangle ABE:

sin(B) = h' / c => h' = c * sin(B)

In right triangle ACE:

sin(C) = h' / b => h' = b * sin(C)

Since both expressions equal h', we can equate them:

c * sin(B) = b * sin(C)

Dividing both sides by sin(B) * sin(C), we get:

c / sin(C) = b / sin(B)

Combining the two results:

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

This completes the proof of the Law of Sines.

Law of Sines: Examples in Code (Python)

import math

def solve_asa(angle_a, angle_c, side_b):
    """Solves a triangle given Angle-Side-Angle (ASA)."""
    angle_b = 180 - angle_a - angle_c
    if angle_b <= 0:
        return "No triangle exists"

    # Convert angles to radians
    angle_a_rad = math.radians(angle_a)
    angle_b_rad = math.radians(angle_b)
    angle_c_rad = math.radians(angle_c)

    # Law of Sines
    side_a = (side_b * math.sin(angle_a_rad)) / math.sin(angle_b_rad)
    side_c = (side_b * math.sin(angle_c_rad)) / math.sin(angle_b_rad)

    return {
        "angle_b": angle_b,
        "side_a": side_a,
        "side_c": side_c
    }

def solve_aas(angle_a, angle_b, side_a):
    """Solves a triangle given Angle-Angle-Side (AAS)."""
    angle_c = 180 - angle_a - angle_b
    if angle_c <= 0:
        return "No triangle exists"

    # Convert angles to radians
    angle_a_rad = math.radians(angle_a)
    angle_b_rad = math.radians(angle_b)
    angle_c_rad = math.radians(angle_c)

    # Law of Sines
    side_b = (side_a * math.sin(angle_b_rad)) / math.sin(angle_a_rad)
    side_c = (side_a * math.sin(angle_c_rad)) / math.sin(angle_a_rad)

    return {
        "angle_c": angle_c,
        "side_b": side_b,
        "side_c": side_c
    }

def solve_ssa(angle_a, side_a, side_b):
    """Solves a triangle given Side-Side-Angle (SSA) - Ambiguous Case."""
    # Convert angle to radians
    angle_a_rad = math.radians(angle_a)

    # Check for the number of possible triangles
    if side_a < side_b * math.sin(angle_a_rad):
        return "No triangle exists"
    elif side_a == side_b * math.sin(angle_a_rad):
        # One right triangle
        angle_b = 90
        angle_c = 180 - angle_a - angle_b
        side_c = (side_a * math.sin(math.radians(angle_c))) / math.sin(angle_a_rad)
        return {
            "angle_b": angle_b,
            "angle_c": angle_c,
            "side_c": side_c
        }
    else:
        # One or two triangles
        sin_b = (side_b * math.sin(angle_a_rad)) / side_a
        angle_b1_rad = math.asin(sin_b)
        angle_b1 = math.degrees(angle_b1_rad)
        angle_b2 = 180 - angle_b1

        # Check if the angles are valid
        angle_c1 = 180 - angle_a - angle_b1
        angle_c2 = 180 - angle_a - angle_b2

        solutions = []

        if angle_c1 > 0:
            side_c1 = (side_a * math.sin(math.radians(angle_c1))) / math.sin(angle_a_rad)
            solutions.append({
                "angle_b": angle_b1,
                "angle_c": angle_c1,
                "side_c": side_c1
            })
        if angle_c2 > 0:
            side_c2 = (side_a * math.sin(math.radians(angle_c2))) / math.sin(angle_a_rad)
            solutions.append({
                "angle_b": angle_b2,
                "angle_c": angle_c2,
                "side_c": side_c2
            })

        if len(solutions) == 0:
            return "No triangle exists"
        elif len(solutions) == 1:
            return solutions[0]
        else:
            return solutions  # Return both possible triangles

# Examples:
print("ASA Example:", solve_asa(30, 70, 10))
print("AAS Example:", solve_aas(40, 60, 8))
print("SSA Example (Ambiguous):", solve_ssa(30, 6, 10))
print("SSA Example (No Triangle):", solve_ssa(30, 3, 10))
print("SSA Example (One Triangle):", solve_ssa(30, 12, 10))

This Python code provides functions to solve triangles given ASA, AAS, and SSA (ambiguous case). It handles the logic for determining the number of possible triangles in the SSA case and returns all possible solutions. The examples demonstrate how to use these functions. Remember to handle the returned values carefully, especially for SSA, which might return a single dictionary, a list of dictionaries, or a "No triangle exists" string. This code provides a practical way to implement the Law of Sines.

Conclusion

The Law of Sines is an essential tool for solving triangles, particularly when dealing with ASA, AAS, and SSA cases. Understanding its applications, especially the ambiguous case, is crucial for accurate problem-solving in various fields. By mastering the Law of Sines and being mindful of potential pitfalls, you can confidently tackle a wide range of triangle-related problems, from academic exercises to real-world applications. Remember to visualize the problem, check for ambiguous cases, and apply the law carefully to arrive at the correct solution.