Given two lines with direction vectors and , use the Law of Cosines to find the acute angle between the lines.
Table of contents
- 0. Review of College Algebra4h 45m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
7. Non-Right Triangles
Law of Cosines
Multiple Choice
Solve the triangle: a=5, b=15, c=13.

A
A=18.9°,B=103.8°,C=57.3°
B
A=13.8°,B=108.9°,C=57.3°
C
A=13.8°,B=103.8°,C=62.4°
D
A=18.9°,B=98.7°,C=62.4°
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Verified step by step guidance1
Identify the given sides of the triangle: a = 5, b = 15, c = 13.
Use the Law of Cosines to find one of the angles. For example, to find angle A, use the formula: \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \). Substitute the values: \( \cos A = \frac{15^2 + 13^2 - 5^2}{2 \times 15 \times 13} \).
Calculate \( \cos A \) using the substituted values to find the cosine of angle A.
Use the inverse cosine function to find angle A: \( A = \cos^{-1}(\cos A) \).
Repeat the process using the Law of Cosines to find angles B and C, using the respective formulas: \( \cos B = \frac{a^2 + c^2 - b^2}{2ac} \) and \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \).
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