Law Of Sines And Cosines Word Problems Worksheet With Answers

The law of sines and the law of cosines are essential tools for solving triangles, particularly when dealing with non-right triangles. Mastering these laws unlocks the ability to calculate unknown angles and sides in various real-world scenarios, from navigation to construction. Let's delve into understanding these laws and how to apply them effectively through solving word problems.

Understanding the Laws

The Law of Sines establishes a relationship between the sides of a triangle and the sines of their opposite angles. It states that for any triangle ABC, with sides a, b, and c opposite to angles A, B, and C respectively:

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

This law is particularly useful when you know:

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

The Law of Cosines, on the other hand, relates the sides of a triangle to the cosine of one of its angles. It's given by the following formulas:

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

The Law of Cosines is your go-to when you know:

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

Strategies for Tackling Word Problems

Before diving into specific examples, let’s establish a general strategy for solving word problems involving the Law of Sines and Law of Cosines:

1. Read Carefully and Visualize: Understand the problem statement thoroughly. Draw a diagram of the triangle described in the problem. Label all known sides and angles.
2. Identify the Given Information: Determine which sides and angles are given. Decide whether you have AAS, ASA, SSA, SSS, or SAS. This will dictate which law to use.
3. Choose the Appropriate Law: Select either the Law of Sines or the Law of Cosines based on the given information.
4. Set Up the Equation: Substitute the known values into the appropriate formula.
5. Solve for the Unknown: Use algebraic techniques to solve for the unknown side or angle.
6. Check Your Answer: Ensure your answer is reasonable within the context of the problem. For example, the largest angle should be opposite the longest side. Also, the sum of angles in any triangle must equal 180 degrees.
7. Units: Always include the correct units in your final answer (e.g., meters, degrees, etc.).

Law of Sines Word Problems: Examples and Solutions

Problem 1:

A surveyor needs to determine the distance across a river. From a point A on one bank, they measure an angle of 68° to a point C on the opposite bank. They then walk 120 meters along the riverbank to point B and measure the angle to point C as 49°. What is the distance across the river (the length of AC)?

Solution:

1. Visualize: Draw a triangle ABC with AB = 120 meters, angle A = 68°, and angle B = 49°. The distance across the river is the length of side AC, which we'll call b.

2. Identify: We have ASA (Angle-Side-Angle).

3. Choose Law: Law of Sines.

4. Set Up: First, find angle C: C = 180° - 68° - 49° = 63°. Now, use the Law of Sines:

   • b / sin(B) = a / sin(A) (We need to find a first to use this directly with b)

   • c / sin(C) = a / sin(A) (We know c and C, so let's use this!)

   • 120 / sin(63°) = a / sin(68°)

5. Solve: Solve for a:

   • a = 120 * sin(68°) / sin(63°)
   • a ≈ 125.7 meters

   Now that we know a, we can solve for b using: a / sin(A) = b / sin(B)

   • 125.7 / sin(68°) = b / sin(49°)
   • b = 125.7 * sin(49°) / sin(68°)
   • b ≈ 101.1 meters

6. Check: The angles and side lengths appear reasonable.

Answer: The distance across the river is approximately 101.1 meters.

Problem 2: The Ambiguous Case (SSA)

A boat is sailing due east. At a certain point, the bearing to a lighthouse is N62°E. After the boat travels 25 kilometers, the bearing to the lighthouse is N28°E. If the boat continues on its course, how close will it get to the lighthouse?

Solution:

1. Visualize: This is tricky! Draw a horizontal line representing the boat's eastward path. Mark the initial position A and the later position B, 25 km apart. Draw lines from A and B to the lighthouse C. The bearing N62°E at A means angle CAB = 90° - 62° = 28°. Similarly, the bearing N28°E at B means angle CBA = 180° - (90° - 28°) = 118°. We want to find the shortest distance from the boat's path to the lighthouse, which is a perpendicular line from C to the path – let's call that point D. We need to find the length of CD.

2. Identify: We have SSA (Side-Side-Angle). Side AB = 25 km, angle CAB = 28°, and we can calculate angle ACB.

3. Choose Law: Law of Sines.

4. Set Up: Find angle ACB: C = 180° - 28° - 118° = 34°. Now use the Law of Sines:

   • 25 / sin(34°) = b / sin(118°) (where b is the length of AC)

5. Solve: Solve for b:

   • b = 25 * sin(118°) / sin(34°)
   • b ≈ 39.3 km

   Now, consider the right triangle ACD. We know AC = 39.3 km and angle CAD = 28°. We want to find CD.

   • sin(28°) = CD / AC
   • CD = AC * sin(28°)
   • CD = 39.3 * sin(28°)
   • CD ≈ 18.5 km

6. Check: The values seem reasonable in the context of the problem. The shortest distance must be less than AC.

Answer: The boat will get approximately 18.5 kilometers close to the lighthouse.

Important Note on the Ambiguous Case (SSA): When using the Law of Sines with SSA, there might be zero, one, or two possible triangles. This occurs because the given side opposite the given angle can sometimes swing into two different positions. Always carefully analyze the problem and check for alternative solutions by considering the supplementary angle (180° - the angle you calculated). If the supplementary angle also results in a valid triangle (all angles sum to 180°), then you have two possible solutions.

Law of Cosines Word Problems: Examples and Solutions

Problem 3:

A triangular plot of land has sides of length 420 meters, 350 meters, and 180 meters. Find the largest angle in the plot of land.

Solution:

1. Visualize: Draw a triangle with sides a = 420 m, b = 350 m, and c = 180 m. The largest angle will be opposite the longest side (420 m), so we want to find angle A.

2. Identify: We have SSS (Side-Side-Side).

3. Choose Law: Law of Cosines.

4. Set Up: Use the Law of Cosines formula that solves for angle A:

   • a² = b² + c² - 2bc cos(A)
   • 420² = 350² + 180² - 2 * 350 * 180 * cos(A)

5. Solve: Solve for cos(A):

   • 176400 = 122500 + 32400 - 126000 * cos(A)
   • 176400 = 154900 - 126000 * cos(A)
   • 21500 = -126000 * cos(A)
   • cos(A) = -21500 / 126000
   • cos(A) ≈ -0.1706

   Now, find angle A using the inverse cosine (arccos):

   • A = arccos(-0.1706)
   • A ≈ 99.8°

6. Check: The angle is obtuse (greater than 90°), which is reasonable given the side lengths.

Answer: The largest angle in the plot of land is approximately 99.8 degrees.

Problem 4:

A ship travels 60 km on a bearing of N20°E and then changes course and travels 80 km on a bearing of N80°E. How far is the ship from its starting point?

Solution:

1. Visualize: Draw a diagram. The ship starts at point A, travels 60 km to point B on a bearing of N20°E, and then travels 80 km to point C on a bearing of N80°E. We want to find the distance AC. The angle between the two paths is crucial. The bearing difference between N20°E and N80°E is 60°. However, this isn't directly the angle ABC. Imagine a North line at point B. The angle between that North line and BA (extended back) is 20°. The angle between the North line and BC is 80°. Therefore, the angle ABC = 180° - (80° - 20°) = 180° - 60° = 120°.

2. Identify: We have SAS (Side-Angle-Side). AB = 60 km, BC = 80 km, and angle ABC = 120°.

3. Choose Law: Law of Cosines.

4. Set Up: We want to find AC, which we'll call b. Use the Law of Cosines formula:

   • b² = a² + c² - 2ac cos(B) (where a = 80, c = 60, and B = 120°)
   • b² = 80² + 60² - 2 * 80 * 60 * cos(120°)

5. Solve: Solve for b:

   • b² = 6400 + 3600 - 9600 * (-0.5) (Remember cos(120°) = -0.5)
   • b² = 10000 + 4800
   • b² = 14800
   • b = √14800
   • b ≈ 121.7 km

6. Check: The distance seems reasonable. It should be greater than both 60 km and 80 km.

Answer: The ship is approximately 121.7 kilometers from its starting point.

Practice Makes Perfect

These examples illustrate the application of the Law of Sines and Law of Cosines in solving word problems. The key to success lies in:

• Careful Reading and Visualization: Understanding the problem and drawing accurate diagrams.
• Correctly Identifying the Given Information: Determining whether you have AAS, ASA, SSA, SSS, or SAS.
• Choosing the Right Law: Selecting the Law of Sines or Law of Cosines based on the given information.
• Accurate Calculations: Paying attention to detail and using a calculator correctly.
• Checking for Reasonableness: Ensuring your answer makes sense in the context of the problem.

Work through as many practice problems as possible to solidify your understanding and build confidence in applying these powerful trigonometric tools. Remember to pay close attention to the ambiguous case (SSA) when using the Law of Sines. With practice, you'll be able to solve a wide variety of triangle-related problems.