6.4 The Law of Sines

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6.4.1 Learning Objectives

By the end of this section, you should be able to:

• apply the Law of Sines;
• solve ASA and AAS triangles;
• analyze the ambiguous SSA case;
• determine whether zero, one, or two triangles exist; and
• solve indirect-measurement problems.

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6.4.2 The Law of Sines

For triangle \(ABC\), with sides \(a,b,c\) opposite angles \(A,B,C\),

\[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. \]

Use it when a known angle-side opposite pair is available.

Example 6.11 Solving an AAS Triangle

In a triangle, \(A=42^\circ\), \(B=73^\circ\), and \(a=12\text{ cm}\). Find \(C\), \(b\), and \(c\).

Solution

\[ C=180^\circ-42^\circ-73^\circ=65^\circ. \]

\[ b=\frac{12\sin73^\circ}{\sin42^\circ}\approx17.15,\qquad c=\frac{12\sin65^\circ}{\sin42^\circ}\approx16.25. \]

\[ \boxed{C=65^\circ,\quad b\approx17.15\text{ cm},\quad c\approx16.25\text{ cm}} \]

Exercise 6.11 Given \(A=35^\circ\), \(B=80^\circ\), and \(a=9\), find \(C\).

Show answer

Answer: \(\boxed{C=65^\circ}\)

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6.4.3 The Ambiguous SSA Case

For SSA information, the equation

\[ \sin B=\frac{b\sin A}{a} \]

may produce no triangle, one triangle, or two triangles because \(\sin B=\sin(180^\circ-B)\). Test the supplementary angle and ensure the angle sum remains below \(180^\circ\).

Example 6.12 Finding Two Possible Triangles

Given \(A=30^\circ\), \(a=10\), and \(b=14\), determine the possible values of \(B\).

Solution

\[ \sin B=\frac{14\sin30^\circ}{10}=0.7. \]

\[ B_1\approx44.43^\circ,\qquad B_2=180^\circ-44.43^\circ\approx135.57^\circ. \]

Both satisfy \(A+B<180^\circ\), so

\[ \boxed{B\approx44.43^\circ\text{ or }135.57^\circ}. \]

Exercise 6.12 Given \(A=40^\circ\), \(a=5\), and \(b=10\), determine whether a triangle exists.

Show answer

Answer: \(\sin B=10\sin40^\circ/5\approx1.286>1\), so \(\boxed{\text{no triangle exists}}\).

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6.4.4 Applications

Law of Sines problems often involve triangulation. Draw and label the triangle, identify an opposite pair, solve, and interpret the requested length or direction.

Example 6.13 Finding an Indirect Distance

Two observation points \(A\) and \(B\) are \(120\text{ m}\) apart. A landmark \(C\) forms angles \(A=48^\circ\) and \(B=67^\circ\). Find \(AC\).

Solution

\[ C=180^\circ-48^\circ-67^\circ=65^\circ. \]

Since \(AB=120\) is opposite \(C\), and \(AC=b\) is opposite \(B\),

\[ b=\frac{120\sin67^\circ}{\sin65^\circ}\approx121.88. \]

\[ \boxed{AC\approx121.88\text{ m}} \]

Exercise 6.13 In a triangle, \(A=52^\circ\), \(C=71^\circ\), and \(c=30\text{ m}\). Find \(a\).

Show answer

Answer: \(\boxed{a=30\sin52^\circ/\sin71^\circ\approx25.00\text{ m}}\)

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6.4.5 Conceptual Takeaways

• The Law of Sines links each side with its opposite angle.
• ASA and AAS information determines one triangle.
• SSA information may be ambiguous.
• Supplementary angles have equal sine values.
• A labelled diagram prevents mismatched side-angle pairs.

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6.4.6 Skills You Should Be Able to Do

• Set up correct Law of Sines proportions.
• Solve ASA and AAS triangles.
• Analyze all SSA possibilities.
• Reject impossible triangles.
• Apply triangulation to indirect measurement.

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6.4.7 Practice Problems with Solutions

1. Given \(A=50^\circ\), \(B=60^\circ\), and \(a=8\), find \(C\).

   Show Solution

   \[ C=180^\circ-50^\circ-60^\circ=\boxed{70^\circ}. \]

2. For Problem 1, find \(b\).

   Show Solution

   \[ b=\frac{8\sin60^\circ}{\sin50^\circ}\approx\boxed{9.04}. \]

3. Given \(A=35^\circ\), \(a=12\), and \(b=8\), find \(B\).

   Show Solution

   \[ \sin B=\frac{8\sin35^\circ}{12}\approx0.3824. \]

   Thus \(B\approx22.48^\circ\). Its supplement cannot work because \(35^\circ+157.52^\circ>180^\circ\).

   \[ \boxed{B\approx22.48^\circ} \]

4. Given \(A=25^\circ\), \(a=7\), and \(b=10\), find all possible values of \(B\).

   Show Solution

   \[ \sin B=\frac{10\sin25^\circ}{7}\approx0.6037. \]

   Therefore, \(B_1\approx37.14^\circ\) and \(B_2\approx142.86^\circ\). Both form valid angle sums with \(A\).

   \[ \boxed{B\approx37.14^\circ\text{ or }142.86^\circ} \]

5. Determine whether a triangle exists if \(A=65^\circ\), \(a=6\), and \(b=9\).

   Show Solution

   \[ \sin B=\frac{9\sin65^\circ}{6}\approx1.359>1. \]

   Thus, \(\boxed{\text{no triangle exists}}\).

6. Two points are \(200\text{ m}\) apart. Their angles to a third point are \(54^\circ\) and \(72^\circ\). Find the distance from the first point to the third.

   Show Solution

   The third angle is \(54^\circ\). If the requested side is opposite \(72^\circ\),

   \[ d=\frac{200\sin72^\circ}{\sin54^\circ}\approx\boxed{235.11\text{ m}}. \]

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