Law Of Sines And Cosines Word Problems Worksheet Answers

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planetorganic

Nov 30, 2025 · 11 min read

Law Of Sines And Cosines Word Problems Worksheet Answers
Law Of Sines And Cosines Word Problems Worksheet Answers

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    The laws of sines and cosines are powerful tools that allow us to solve triangles, especially when we lack the nice, neat conditions of right triangles. When applied skillfully, these laws open the door to solving a wide array of real-world problems involving angles and distances. Mastering these laws begins with understanding the formulas themselves and then extends to recognizing the types of problems where each law is most applicable.

    Understanding the Laws: A Quick Review

    Before diving into word problems, let’s briefly recap the Law of Sines and the Law of Cosines:

    • Law of Sines: In any triangle ABC, with sides a, b, and c opposite to angles A, B, and C, respectively, the following relationship holds:

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

      The Law of Sines is particularly useful when you know:

      • Two angles and one side (AAS or ASA)
      • Two sides and an angle opposite one of them (SSA) – beware of the ambiguous case!
    • Law of Cosines: In any triangle ABC, the following relationships hold:

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

      The Law of Cosines is particularly useful when you know:

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

    Identifying the Right Law for the Job

    The key to solving word problems involving triangles is recognizing which law to apply. Ask yourself these questions:

    1. What information am I given? Carefully list the known angles and sides.
    2. What am I trying to find? Identify the unknown angle or side.
    3. Does the given information fit the conditions for the Law of Sines (AAS, ASA, SSA) or the Law of Cosines (SSS, SAS)?

    Let's work through a variety of word problems to illustrate these concepts.

    Example Word Problems and Solutions

    Problem 1: The Leaning Tower of Pisa

    The Leaning Tower of Pisa is inclined 5.5° from the vertical. A point 80 meters from the base of the tower is sighted, and the angle of elevation to the top of the tower is 29°. Approximate the height of the tower.

    Solution:

    1. Draw a Diagram: Always start with a clear diagram. Draw a triangle representing the tower, the ground, and the line of sight. Mark the known angles and distances.

    2. Identify the Triangle: We have a triangle that isn't a right triangle. We know one side (80 meters) and we can find two angles.

    3. Calculate the Angles:

      • The angle at the base of the tower is 90° + 5.5° = 95.5°
      • The angle at the top of the triangle (opposite the 80-meter side) is 180° - 95.5° - 29° = 55.5°
    4. Apply the Law of Sines: We want to find the height of the tower (h), which is opposite the 29° angle. Using the Law of Sines:

      h / sin(29°) = 80 / sin(55.5°)
      
    5. Solve for h:

      h = 80 * sin(29°) / sin(55.5°)
      h ≈ 46.1 meters
      

      Therefore, the height of the Leaning Tower of Pisa is approximately 46.1 meters.

    Problem 2: Navigation

    A boat leaves port and sails 12 nautical miles on a bearing of N 40° E. Then, the boat turns and sails 8 nautical miles on a bearing of E 20° S. How far is the boat from port? What is the bearing from port to the boat?

    Solution:

    1. Draw a Diagram: This is crucial for navigation problems. Draw a compass rose, then sketch the boat's path.

    2. Identify the Triangle: We have a triangle where we know two sides (12 and 8 nautical miles) and the included angle. We need to find the third side and an angle to determine the final bearing.

    3. Calculate the Included Angle: The angle between the two legs of the journey is 90° - 40° + 90° - 20° = 120°. This isn't quite right. Consider the internal angle of the triangle formed by the two legs of the journey. The first leg is N 40° E, and the second leg is E 20° S. The angle inside the triangle between these two legs is 180 - (50 + 70) = 120 degrees.

    4. Apply the Law of Cosines: Let d be the distance from the port to the boat.

      d² = 12² + 8² - 2 * 12 * 8 * cos(120°)
      d² = 144 + 64 - 192 * (-0.5)
      d² = 208 + 96
      d² = 304
      d = √304 ≈ 17.4 nautical miles
      
    5. Find the Bearing: We need to find the angle between the north direction and the line connecting the port to the boat. Let's call the angle at the port θ. Use the Law of Sines to find θ:

      sin(θ) / 8 = sin(120°) / 17.4
      sin(θ) = (8 * sin(120°)) / 17.4
      sin(θ) ≈ 0.398
      θ ≈ arcsin(0.398) ≈ 23.5°
      

      This angle is east of the initial northward heading. Therefore, the bearing from the port to the boat is approximately N 23.5° E.

    Problem 3: Bridge Construction

    Engineers are building a bridge across a river. From one side of the river, they measure the angle of elevation to the top of a tree on the opposite bank to be 32°. They then walk 50 meters along the riverbank and measure the angle of elevation to the same treetop to be 25°. How tall is the tree?

    Solution:

    1. Draw a Diagram: Draw two triangles, one nested inside the other, sharing the treetop as a common vertex.

    2. Identify the Triangles: We need to create a triangle where we can use the Law of Sines or Cosines. We can create a triangle with vertices at the treetop, the first observation point, and the second observation point.

    3. Calculate the Angles:

      • The angle at the first observation point is 32°.
      • The angle at the second observation point is the supplement of the angle formed by the 25° elevation and the horizontal line (180° - 25°). However, we need the angle inside the triangle. Let the angle formed by walking along the bank be alpha. Alpha is 180 - 32 = 148. Let beta be 180 - 25 = 155. The angle we need is the supplement of the difference between these two, 180 - (155-148) = 173. This is the WRONG APPROACH.

      We need to find the angle at the treetop. Let's call it γ. Consider the large triangle (from the first observation point to the treetop to the base of the tree). The angle at the base of the tree is 90. The angle at the first observation point is 32. Therefore, the angle at the top is 180 - 90 - 32 = 58. Next, consider the small triangle (from the second observation point to the treetop to the base of the tree). The angle at the second observation point is 25. Therefore the angle at the top is 180 - 90 - 25 = 65. The angle gamma is therefore 65 - 58 = 7.

      The angle opposite the 50-meter side is 7 degrees.

      Let's find the distance from the second observation point to the top of the tree. Call this D.

    4. Apply the Law of Sines: Find the length of the side connecting the second observation point to the treetop.

      50 / sin(7) = D / sin(32)
      D = 50 * sin(32) / sin(7)
      D = 216.36
      
    5. Find the height: Use trigonometry and the angle of 25 degrees.

      sin(25) = h / D
      h = sin(25) * D
      h = sin(25) * 216.36
      h = 91.39
      

      Therefore the height of the tree is 91.39 meters.

    Problem 4: Surveying Land

    A surveyor wants to determine the distance between two points, A and B, on opposite sides of a lake. From a point C, the surveyor measures the distance to point A as 250 meters and the distance to point B as 320 meters. The angle ACB is measured to be 58°. Find the distance between points A and B.

    Solution:

    1. Draw a Diagram: Draw a triangle with vertices A, B, and C. Label the known sides and angle.

    2. Identify the Triangle: We know two sides (250 m and 320 m) and the included angle (58°).

    3. Apply the Law of Cosines: Let c be the distance between A and B.

      c² = 250² + 320² - 2 * 250 * 320 * cos(58°)
      c² = 62500 + 102400 - 160000 * cos(58°)
      c² ≈ 164900 - 84784
      c² ≈ 80116
      c ≈ √80116 ≈ 283.05 meters
      

      Therefore, the distance between points A and B is approximately 283.05 meters.

    Problem 5: Airplane Flight

    An airplane flies on a bearing of 70° for a distance of 500 miles. It then changes direction and flies on a bearing of 115° for 350 miles. How far is the airplane from its starting point, and what is the bearing from the starting point to the airplane?

    Solution:

    1. Draw a Diagram: Draw a compass rose and sketch the airplane's path.

    2. Identify the Triangle: We have a triangle where we know two sides and the included angle.

    3. Calculate the Included Angle: The angle between the two flight paths is the difference between the bearings. The second leg is 115 - 70 = 45 + 180 = 225. The angle inside the triangle formed by the two legs of the journey is 180 - 45 = 135.

    4. Apply the Law of Cosines: Let d be the distance from the starting point to the airplane.

      d² = 500² + 350² - 2 * 500 * 350 * cos(135°)
      d² = 250000 + 122500 - 350000 * (-√2/2)
      d² ≈ 372500 + 247487
      d² ≈ 619987
      d ≈ √619987 ≈ 787.39 miles
      
    5. Find the Bearing: Let's call the angle at the starting point θ. Use the Law of Sines to find θ:

      sin(θ) / 350 = sin(135°) / 787.39
      sin(θ) = (350 * sin(135°)) / 787.39
      sin(θ) ≈ 0.314
      θ ≈ arcsin(0.314) ≈ 18.3°
      

      This angle is added to the initial bearing of 70°. Therefore, the bearing from the starting point to the airplane is approximately 70° + 18.3° = 88.3°. So it is a bearing of approximately 88.3 degrees.

    Problem 6: Baseball Diamond

    A baseball diamond is a square with sides of 90 feet. The pitcher's mound is 60.5 feet from home plate. How far is the pitcher's mound from first base?

    Solution:

    1. Draw a Diagram: Draw the square baseball diamond and the location of the pitcher's mound.

    2. Identify the Triangle: We can form a triangle with vertices at home plate, first base, and the pitcher's mound. We know two sides (90 feet and 60.5 feet) and the included angle (45° since the line from home to first bisects the right angle at home plate).

    3. Apply the Law of Cosines: Let d be the distance from the pitcher's mound to first base.

      d² = 90² + 60.5² - 2 * 90 * 60.5 * cos(45°)
      d² = 8100 + 3660.25 - 10890 * cos(45°)
      d² ≈ 11760.25 - 7699.6
      d² ≈ 4060.65
      d ≈ √4060.65 ≈ 63.72 feet
      

      Therefore, the distance from the pitcher's mound to first base is approximately 63.72 feet.

    Problem 7: A Triangular Garden

    A triangular garden has sides of length 15 feet, 18 feet, and 22 feet. Find the largest angle in the garden.

    Solution:

    1. Draw a Diagram: Draw a triangle with sides labeled 15, 18, and 22.

    2. Identify the Triangle: We know all three sides (SSS). The largest angle will be opposite the longest side (22 feet).

    3. Apply the Law of Cosines: Let C be the angle opposite the side of length 22.

      22² = 15² + 18² - 2 * 15 * 18 * cos(C)
      484 = 225 + 324 - 540 * cos(C)
      484 = 549 - 540 * cos(C)
      -65 = -540 * cos(C)
      cos(C) = 65 / 540 ≈ 0.1204
      C ≈ arccos(0.1204) ≈ 83.1°
      

      Therefore, the largest angle in the garden is approximately 83.1 degrees.

    Tips and Tricks for Success

    • Draw Diagrams: This cannot be stressed enough. A visual representation helps immensely.
    • Label Everything: Clearly label all known and unknown sides and angles.
    • Choose the Right Law: Carefully consider whether the Law of Sines or the Law of Cosines is more appropriate.
    • Watch for the Ambiguous Case (SSA): When using the Law of Sines with SSA, be aware that there might be zero, one, or two possible triangles.
    • Use Units: Keep track of units (meters, feet, miles, etc.) throughout the problem.
    • Check Your Answers: Does the answer make sense in the context of the problem? For example, the longest side should be opposite the largest angle.

    Common Mistakes to Avoid

    • Incorrectly Applying the Laws: Double-check that you're using the correct formulas and substituting values correctly.
    • Forgetting the Ambiguous Case: Always consider the possibility of two solutions when using the Law of Sines with SSA.
    • Rounding Errors: Avoid rounding intermediate calculations too early, as this can lead to significant errors in the final answer.
    • Misinterpreting Bearings: Be careful when interpreting bearings; ensure you understand the reference direction (north or south).
    • Not Drawing a Diagram: This is a recipe for disaster.

    Conclusion

    Mastering the Law of Sines and the Law of Cosines opens up a world of possibilities for solving real-world problems involving triangles. By carefully drawing diagrams, identifying the given information, choosing the appropriate law, and avoiding common mistakes, you can confidently tackle even the most challenging word problems. Practice is key, so work through a variety of examples to solidify your understanding. Remember to always think critically about the problem and ensure your answers make sense in context. Good luck!

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