Law Of Cosines Practice Problems

Mastering the Law of Cosines: Practice Problems and Solutions

The Law of Cosines is a fundamental theorem in trigonometry, extending the Pythagorean theorem to non-right-angled triangles. It provides a powerful tool for solving problems involving the lengths of sides and the angles of any triangle, whether acute, obtuse, or right-angled. This comprehensive guide will equip you with a thorough understanding of the Law of Cosines, providing a range of practice problems with detailed solutions to solidify your grasp of this crucial concept. Understanding and applying the Law of Cosines is essential for anyone studying geometry, trigonometry, or related fields.

Understanding the Law of Cosines

The Law of Cosines states the relationship between the lengths of the sides of a triangle and the cosine of one of its angles. For any triangle with sides a, b, and c, and angles A, B, and C (opposite to sides a, b, and c respectively), the Law of Cosines can be expressed in three forms:

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

Notice that if the angle is 90° (a right angle), cos 90° = 0, and the formula simplifies to the Pythagorean theorem (a² = b² + c²). This highlights the Law of Cosines' role as a generalization of the Pythagorean theorem.

Practice Problems: Beginner Level

Let's start with some basic problems to build your confidence. Remember to always draw a diagram to visualize the triangle and label its sides and angles correctly.

Problem 1:

A triangle has sides a = 5 cm, b = 7 cm, and angle C = 60°. Find the length of side c.

Solution:

We use the Law of Cosines formula: c² = a² + b² - 2ab cos C

Substituting the given values:

c² = 5² + 7² - 2(5)(7) cos 60° c² = 25 + 49 - 70 (0.5) c² = 74 - 35 c² = 39 c = √39 ≈ 6.24 cm

Problem 2:

A triangle has sides a = 8 m and b = 10 m, and angle C = 120°. Find the length of side c.

Solution:

Using the Law of Cosines: c² = a² + b² - 2ab cos C

Substituting the values:

c² = 8² + 10² - 2(8)(10) cos 120° c² = 64 + 100 - 160 (-0.5) c² = 164 + 80 c² = 244 c = √244 ≈ 15.62 m

Problem 3:

Two sides of a triangle measure 12 cm and 15 cm, and the angle between them is 45°. Find the length of the third side.

Solution:

Let a = 12 cm, b = 15 cm, and C = 45°. We use the Law of Cosines to find c:

c² = a² + b² - 2ab cos C c² = 12² + 15² - 2(12)(15) cos 45° c² = 144 + 225 - 360 (√2/2) c² = 369 - 180√2 c² ≈ 369 - 254.56 c² ≈ 114.44 c ≈ √114.44 ≈ 10.7 cm

Practice Problems: Intermediate Level

These problems involve slightly more complex scenarios and require a deeper understanding of the Law of Cosines.

Problem 4:

Find the largest angle in a triangle with sides of length 6, 8, and 10.

Solution:

The largest angle is opposite the longest side. Let a = 6, b = 8, and c = 10. We'll use the Law of Cosines to find angle C:

c² = a² + b² - 2ab cos C 10² = 6² + 8² - 2(6)(8) cos C 100 = 36 + 64 - 96 cos C 0 = -96 cos C cos C = 0 C = 90°

Therefore, the largest angle is 90°. This triangle is a right-angled triangle.

Problem 5:

A triangle has sides of length 7, 9, and 12. Find all three angles.

Solution:

Let a = 7, b = 9, and c = 12. We'll use the Law of Cosines to find each angle:

• Finding angle A: a² = b² + c² - 2bc cos A 7² = 9² + 12² - 2(9)(12) cos A 49 = 81 + 144 - 216 cos A 216 cos A = 176 cos A = 176/216 ≈ 0.815 A = arccos(0.815) ≈ 35.4°

• Finding angle B: b² = a² + c² - 2ac cos B 9² = 7² + 12² - 2(7)(12) cos B 81 = 49 + 144 - 168 cos B 168 cos B = 112 cos B = 112/168 ≈ 0.667 B = arccos(0.667) ≈ 48.2°

• Finding angle C: c² = a² + b² - 2ab cos C 12² = 7² + 9² - 2(7)(9) cos C 144 = 49 + 81 - 126 cos C 126 cos C = -14 cos C = -14/126 ≈ -0.111 C = arccos(-0.111) ≈ 96.4°

Note: The sum of the angles should approximately equal 180° (35.4° + 48.2° + 96.4° ≈ 180°). Slight discrepancies might arise due to rounding errors.

Practice Problems: Advanced Level

These problems require a deeper understanding of trigonometric concepts and problem-solving skills.

Problem 6:

Two ships leave a port at the same time. One ship travels at 15 km/h on a bearing of 060°. The other ship travels at 20 km/h on a bearing of 135°. After 2 hours, what is the distance between the two ships?

Solution:

First, calculate the distance traveled by each ship:

• Ship 1: Distance = speed × time = 15 km/h × 2 h = 30 km
• Ship 2: Distance = speed × time = 20 km/h × 2 h = 40 km

The angle between their paths is 135° - 60° = 75°. We can now use the Law of Cosines to find the distance (d) between the ships:

d² = 30² + 40² - 2(30)(40) cos 75° d² = 900 + 1600 - 2400 cos 75° d² ≈ 2500 - 621.17 d² ≈ 1878.83 d ≈ √1878.83 ≈ 43.35 km

Problem 7:

A surveyor wants to measure the distance across a river. He stands at point A and sights a point B on the opposite bank. Then, he walks 100 meters to point C, where the angle ACB is 70° and the angle BAC is 40°. Find the width of the river (distance AB).

Solution:

This problem requires using the Law of Sines in conjunction with the Law of Cosines. First, we find angle ABC:

Angle ABC = 180° - 70° - 40° = 70°

Since angles ABC and ACB are equal, this is an isosceles triangle with AB = AC. Now we can use the Law of Sines:

AB/sin 70° = 100 m/sin 40° AB = 100 m * (sin 70° / sin 40°) AB ≈ 146.19 m

The width of the river is approximately 146.19 meters.

Frequently Asked Questions (FAQ)

Q: When should I use the Law of Cosines instead of the Law of Sines?

A: You use the Law of Cosines when you know:

• Three sides of a triangle (SSS) and you need to find an angle.
• Two sides and the included angle (SAS) and you need to find the third side.

The Law of Sines is used when you know:

• Two angles and a side (AAS or ASA)
• Two sides and a non-included angle (SSA - ambiguous case).

Q: Can the Law of Cosines be used for right-angled triangles?

A: Yes, it can. However, for right-angled triangles, the Pythagorean theorem is simpler and more efficient. The Law of Cosines reduces to the Pythagorean theorem when the angle is 90°.

Q: What if I get a negative value when solving for cos θ?

A: A negative value for cos θ indicates that the angle θ is obtuse (greater than 90°). Use the inverse cosine function (arccos) to find the angle's value.

Conclusion

The Law of Cosines is a versatile tool for solving various triangle problems. Mastering its application requires consistent practice and understanding its relationship with the Pythagorean theorem and the Law of Sines. By working through these practice problems, you should have a much stronger understanding of this fundamental trigonometric concept and feel confident in tackling more advanced problems in geometry and trigonometry. Remember that drawing a diagram is crucial for visualizing the problem and avoiding common mistakes. Keep practicing, and you’ll become proficient in solving any triangle problem thrown your way!