Law of Sines and Law of Cosines

Overview

The Law of Sines and the Law of Cosines are powerful tools for solving oblique (non-right) triangles. These laws are particularly useful when we know certain angles and sides of a triangle and need to find missing sides or angles.

Law of Sines

The Law of Sines relates the sides of a triangle to the sines of its angles. It is expressed as:

\( \frac{a}{sin(A)} = \frac{b}{sin(B)} = \frac{c}{sin(C)} \)

Where \(a\), \(b\), and \(c\) are the sides of the triangle, and \(A\), \(B\), and \(C\) are the opposite angles. This law is useful when we know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA).

Law of Cosines

The Law of Cosines relates the sides of a triangle to the cosine of one of its angles. It is expressed as:

\( c² = a² + b² - 2ab \cdot cos(C) \)

Where \(a\), \(b\), and \(c\) are the sides, and \(C\) is the angle opposite side \(c\). The Law of Cosines is helpful when we know two sides and the included angle (SAS) or three sides (SSS).

Key Concepts

  • Law of Sines: Used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA).
  • Law of Cosines: Used when you know two sides and the included angle (SAS) or all three sides (SSS).
  • Solving for Missing Angles: The Law of Sines can be used to find missing angles when the opposite sides are known, while the Law of Cosines helps in cases involving missing sides or angles where the two sides and the included angle are known.
  • Ambiguity in SSA: When using the Law of Sines in SSA cases, be aware of the possible ambiguous solutions, which may result in no solution, one solution, or two solutions.

Practice Problems

  1. In triangle ABC, \(A = 30^\circ\), \(B = 45^\circ\), and \(c = 10\). Use the Law of Sines to find side \(a\).
    Solution

    First, calculate angle \(C\): \(C = 180^\circ - A - B = 180^\circ - 30^\circ - 45^\circ = 105^\circ\).

    Now use the Law of Sines to find side \(a\):

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

    Solving for \(a\): \(a = \frac{10 \cdot \sin(30^\circ)}{\sin(105^\circ)} \approx 5.89\).

  2. In triangle ABC, \(a = 7\), \(b = 10\), and \(C = 60^\circ\). Use the Law of Cosines to find side \(c\).
    Solution

    Using the Law of Cosines:

    c² = a² + b² - 2ab * cos(C)

    Substitute the given values:

    c² = 7² + 10² - 2(7)(10) * cos(60^\circ)

    Calculate the right-hand side:

    c² = 49 + 100 - 140 * 0.5 = 49 + 100 - 70 = 79

    Taking the square root:

    c = \sqrt{79} \approx 8.89

  3. In triangle XYZ, \(X = 60^\circ\), \(Y = 40^\circ\), and \(z = 12\). Use the Law of Sines to find angle \(Z\).
    Solution

    First, find angle \(Z\):

    Z = 180^\circ - X - Y = 180^\circ - 60^\circ - 40^\circ = 80^\circ
  4. In triangle PQR, side \(p = 5\), side \(q = 7\), and angle \(R = 90^\circ\). Use the Law of Cosines to find side \(r\).
    Solution

    Since angle \(R = 90^\circ\), the Law of Cosines simplifies to the Pythagorean Theorem:

    r² = p² + q²

    Substitute the values:

    r² = 5² + 7² = 25 + 49 = 74

    Taking the square root:

    r = \sqrt{74} \approx 8.60
  5. In triangle DEF, angle \(D = 50^\circ\), angle \(E = 60^\circ\), and side \(d = 8\). Use the Law of Sines to find side \(e\).
    Solution

    First, calculate angle \(F\):

    F = 180^\circ - D - E = 180^\circ - 50^\circ - 60^\circ = 70^\circ

    Now use the Law of Sines:

    e / sin(E) = d / sin(D)

    Solving for \(e\):

    e = \frac{8 \cdot \sin(60^\circ)}{\sin(50^\circ)} \approx 9.29
  6. In a triangle, side \(a = 9\), side \(b = 6\), and angle \(C = 75^\circ\). Use the Law of Cosines to find angle \(A\).
    Solution

    Using the Law of Cosines to find side \(c\) first:

    c² = a² + b² - 2ab * cos(C)

    Substitute the values:

    c² = 9² + 6² - 2(9)(6) * cos(75^\circ)

    After calculating, find the value for side \(c\). Then use the Law of Sines to find angle \(A\).

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