7.6 Law of Sines, Law of Cosines, and Triangle Applications

7.6 Law of Sines, Law of Cosines, and Triangle Applications

Right-triangle trigonometry applies only when one angle is exactly $90^\circ$. For non-right (oblique) triangles, the Law of Sines and Law of Cosines extend trigonometric relationships to any triangle.

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1. Standard Notation and Triangle Cases

In an oblique triangle $ABC$, the angles are denoted by uppercase letters $A, B, C$, and their opposite side lengths are denoted by lowercase letters $a, b, c$. The sum of angles is always $A + B + C = 180^\circ$.

To solve an oblique triangle (find all missing sides and angles), one must know three measurements, categorized into five standard cases:

1. AAS (Angle-Angle-Side): Law of Sines
2. ASA (Angle-Side-Angle): Law of Sines
3. SSA (Side-Side-Angle): Law of Sines (Ambiguous Case)
4. SAS (Side-Angle-Side): Law of Cosines
5. SSS (Side-Side-Side): Law of Cosines

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2. Law of Sines

Law of Sines: In any triangle $ABC$:

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

The Ambiguous Case (SSA)

When given two sides and a non-included angle (SSA), the given information may determine zero, one, or two triangles. For an acute angle $A$ with given sides $a$ and $b$, calculate the altitude $h = b \sin A$:

If angle $A$ is obtuse ($\ge 90^\circ$):

• If $a \le b$: $0$ Triangles
• If $a > b$: $1$ Triangle

Worked Example 1: SSA Ambiguous Case (Zero Solutions)

Determine the number of possible triangles if $A = 30^\circ$, $b = 12$, and $a = 5$.

Solution:

1. Calculate altitude $h$: $h = b \sin A = 12 \sin(30^\circ) = 12 (0.5) = 6$
2. Compare side $a = 5$ to altitude $h = 6$: $a < h \quad (5 < 6)$
3. Since side $a$ is shorter than the minimum height required to reach the base, no triangle exists ($0$ solutions).

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3. Law of Cosines

Law of Cosines: In any triangle $ABC$:

$$$$

a^2 &= b^2 + c^2 - 2bc \cos A \ b^2 &= a^2 + c^2 - 2ac \cos B \ c^2 &= a^2 + b^2 - 2ab \cos C \end{aligned}$$

To solve for an angle given three sides (SSS), rearrange the formula:

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

Best Practice Tip for SSS: Always solve for the largest angle first (the angle opposite the longest side). Since $\arccos$ returns values in $[0^\circ, 180^\circ]$, $\cos A < 0$ instantly identifies whether the triangle has an obtuse angle!

Worked Example 2: Solving an SAS Triangle

Given a triangle with $a = 7$, $b = 10$, and included angle $C = 60^\circ$, calculate the exact length of side $c$.

Solution:

1. Apply the Law of Cosines for side $c$: $c^2 = a^2 + b^2 - 2ab \cos C$
2. Substitute the given values: $c^2 = 7^2 + 10^2 - 2(7)(10) \cos(60^\circ)$ $c^2 = 49 + 100 - 140(0.5) = 149 - 70 = 79$
3. Take the square root: $c = \sqrt{79}$

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4. Triangle Area Formulas

Side-Angle-Side (SAS) Area Formula

The area $K$ of a triangle with two sides and an included angle is:

$K = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B$

Heron's Formula (SSS Area)

When all three side lengths $a, b, c$ are known, calculate the semi-perimeter $s$:

$s = \frac{a + b + c}{2}$

Then the area $K$ is:

$K = \sqrt{s(s - a)(s - b)(s - c)}$

Worked Example 3: Area via Heron's Formula

Calculate the area of a triangle with side lengths $a = 5$, $b = 6$, and $c = 7$.

Solution:

1. Compute the semi-perimeter $s$: $s = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9$
2. Apply Heron's formula: $K = \sqrt{9(9 - 5)(9 - 6)(9 - 7)} = \sqrt{9(4)(3)(2)} = \sqrt{216}$
3. Simplify the radical: $K = \sqrt{36 \times 6} = 6\sqrt{6}$

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5. CLEP Exam Traps & Common Errors

• Trap 1: Applying Law of Sines to SAS Triangles First
  Trying to use Law of Sines when given SAS leads to an equation with two unknown variables. Use Law of Cosines first to find the third side!
• Trap 2: Ignoring the Second Triangle in SSA
  When solving SSA with $h < a < b$, using inverse sine gives the acute angle $B_1$. You MUST check $B_2 = 180^\circ - B_1$. If $A + B_2 < 180^\circ$, a valid second triangle exists!
• Trap 3: Order of Operations in Law of Cosines
  A classic computational error is writing $a^2 + b^2 - 2ab \cos C$ as $(a^2 + b^2 - 2ab) \cos C$. Multiplication takes precedence: $2ab \cos C$ is a single term subtracted from $a^2 + b^2$.