12.3 Law of Sines (Ambiguous Case), Law of Cosines & Oblique Triangle Area

12.3 Law of Sines (Ambiguous Case), Law of Cosines & Oblique Triangle Area

Analytical Mechanics of Oblique Triangles

An oblique triangle is any planar triangle that contains no $90^\circ$ right angle, classifying it as either acute (all interior angles strictly less than $90^\circ$) or obtuse (one interior angle strictly greater than $90^\circ$). Because right-triangle definitions cannot be applied directly to oblique triangles without dropping an altitude, two fundamental trigonometric theorems govern oblique triangle resolution: the Law of Sines and the Law of Cosines.

In standard Euclidean notation, the vertices of triangle $ABC$ are denoted by uppercase letters $A, B, C$, their corresponding interior angle measures are denoted by the same letters, and the lengths of the sides opposite these angles are denoted by lowercase letters $a, b, c$.

Depending on the given geometric data, triangle resolution falls into five distinct configuration cases:

1. AAS or ASA (Angle-Angle-Side or Angle-Side-Angle): Solved uniquely using the Law of Sines.
2. SSA (Side-Side-Angle): The Ambiguous Case. Because the known angle is not included between the two known sides, this configuration may yield zero, one, or two distinct triangles.
3. SAS (Side-Angle-Side): Solved using the Law of Cosines to find the third side, followed by the Law of Sines or Law of Cosines for remaining angles.
4. SSS (Side-Side-Side): Solved using the Law of Cosines to find the largest angle first, preventing inverse-sine quadrant ambiguity.

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The Law of Sines and the Ambiguous Case (SSA)

For any triangle $ABC$ inscribed in a circle of circumradius $R$, the ratio of each side length to the sine of its opposite angle is constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ The Law of Sines is derived directly by dropping an altitude $h$ from vertex $C$ to side $c$: $h = b\sin A = a\sin B \implies \frac{a}{\sin A} = \frac{b}{\sin B}$

The Ambiguous Case (SSA) Analysis

When given an angle $A$, an adjacent side $b$, and an opposite side $a$, the geometric altitude from vertex $C$ perpendicular to the baseline along which side $c$ lies is: $h = b\sin A$

The existence and uniqueness of the resulting triangle depend strictly on the classification of angle $A$ and the magnitude of side $a$ relative to altitude $h$ and adjacent side $b$:

1. When Angle $A$ is Acute ($A < 90^\circ$):

• Case 1: $a < h$ ($a < b\sin A$): The opposite side $a$ is shorter than the perpendicular altitude $h$. Side $a$ cannot reach the baseline line of sight. No triangle exists (0 solutions).
• Case 2: $a = h$ ($a = b\sin A$): Side $a$ reaches the baseline perpendicularly at exactly one point. Exactly one right triangle exists ($\angle B = 90^\circ$).
• Case 3: $h < a < b$ ($b\sin A < a < b$): Side $a$ is longer than altitude $h$ but shorter than adjacent side $b$. A circle of radius $a$ centered at $C$ cuts the baseline at two distinct positive locations, creating two non-congruent triangles:
  • Triangle 1 features an acute angle $B_1 = \arcsin\left(\frac{b\sin A}{a}\right)$.
  • Triangle 2 features an obtuse angle $B_2 = 180^\circ - B_1$. Both angles are valid because $A + B_2 < A + (180^\circ - A) = 180^\circ$, leaving room for a positive angle $C_2 = 180^\circ - (A + B_2)$. Two distinct triangles exist (2 solutions).
• Case 4: $a \ge b$: Side $a$ is greater than or equal to adjacent side $b$. The circle of radius $a$ centered at $C$ intersects the baseline at only one point in the forward direction (the other intersection lies behind vertex $A$, forming an exterior non-triangle angle). Exactly one unique triangle exists (1 solution).

2. When Angle $A$ is Obtuse ($A \ge 90^\circ$):

• Case 1: $a \le b$: In any obtuse triangle, the side opposite the obtuse angle must be strictly the longest side. If $a \le b$, no triangle can close. No triangle exists (0 solutions).
• Case 2: $a > b$: Side $a$ exceeds adjacent side $b$, closing the triangle uniquely. Exactly one triangle exists (1 solution).

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The Law of Cosines: Formulation and Application

The Law of Cosines generalizes the Pythagorean theorem to arbitrary triangles by introducing a correction term proportional to the cosine of the included angle: $c^2 = a^2 + b^2 - 2ab\cos C$ $b^2 = a^2 + c^2 - 2ac\cos B$ $a^2 = b^2 + c^2 - 2bc\cos A$

When angle $C = 90^\circ$, $\cos 90^\circ = 0$, and the formula reduces to $c^2 = a^2 + b^2$.

Solving the SSS Case: Avoiding Inverse Sine Ambiguity

When all three sides $a, b, c$ are known, solve for the angle opposite the longest side first using the isolated cosine formula: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ This strategy is essential: because the inverse cosine function $\arccos(x)$ maps the domain $[-1, 1]$ monotonically onto the range $[0^\circ, 180^\circ]$:

• If $\cos C > 0$, angle $C$ is uniquely acute.
• If $\cos C = 0$, angle $C = 90^\circ$ is uniquely a right angle.
• If $\cos C < 0$, angle $C$ is uniquely obtuse. Once the largest angle is identified, any remaining unknown angles must be strictly acute, allowing safe evaluation via the Law of Sines without quadrant ambiguity.

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Oblique Triangle Area Formulations

1. The SAS Sine Area Formula

In any triangle with two known side lengths and their included angle, the altitude to one side is $h = b\sin C$. Substituting $h$ into the standard area formula $A = \frac{1}{2}ah$ yields: $\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B$

2. Heron's Formula (SSS)

When all three side lengths $a, b, c$ are known, Heron's formula computes the exact area without calculating altitudes or interior angles: $\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}$ where $s$ is the semi-perimeter: $s = \frac{a + b + c}{2}$

Heron's formula connects to the triangle's inradius $r$ and circumradius $R$ via: $\text{Area} = r \cdot s = \frac{abc}{4R}$

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Worked Exemplar: Resolving the SSA Ambiguous Case

Problem: In triangle $ABC$, side $a = 7$, side $b = 10$, and angle $A = 30^\circ$. Determine the number of possible triangles, and find the complete solution (all angles and side lengths) for each possible triangle.

• Step 1: Calculate the Perpendicular Altitude $h$. $h = b\sin A = 10\sin 30^\circ = 10(0.5) = 5$

• Step 2: Compare Side Lengths to Determine Solution Multiplicity. We compare $a = 7$ to altitude $h = 5$ and adjacent side $b = 10$: $h < a < b \implies 5 < 7 < 10$ Because side $a$ is strictly between the altitude $h$ and adjacent side $b$, the SSA condition yields exactly two distinct triangles ($\triangle ABC_1$ and $\triangle ABC_2$).

• Step 3: Solve Triangle 1 (Acute Angle $B_1$). Apply the Law of Sines: $\frac{\sin B_1}{b} = \frac{\sin A}{a} \implies \sin B_1 = \frac{10\sin 30^\circ}{7} = \frac{5}{7} \approx 0.714286$ The acute angle solution is: $B_1 = \arcsin\left(\frac{5}{7}\right) \approx 45.58^\circ$ Find third angle $C_1$: $C_1 = 180^\circ - (A + B_1) = 180^\circ - (30^\circ + 45.58^\circ) = 104.42^\circ$ Find side $c_1$ using the Law of Sines: $c_1 = \frac{a\sin C_1}{\sin A} = \frac{7\sin(104.42^\circ)}{\sin 30^\circ} = \frac{7(0.9685)}{0.5} \approx 13.56$

• Step 4: Solve Triangle 2 (Obtuse Angle $B_2$). The obtuse supplement of $B_1$ is: $B_2 = 180^\circ - B_1 = 180^\circ - 45.58^\circ = 134.42^\circ$ Check angle sum validity: $A + B_2 = 30^\circ + 134.42^\circ = 164.42^\circ < 180^\circ$ Find third angle $C_2$: $C_2 = 180^\circ - (A + B_2) = 180^\circ - 164.42^\circ = 15.58^\circ$ Find side $c_2$ using the Law of Sines: $c_2 = \frac{a\sin C_2}{\sin A} = \frac{7\sin(15.58^\circ)}{\sin 30^\circ} = \frac{7(0.2686)}{0.5} \approx 3.76$

• Summary of Solutions:

  • Triangle 1: $B_1 \approx 45.58^\circ$, $C_1 \approx 104.42^\circ$, $c_1 \approx 13.56$.
  • Triangle 2: $B_2 \approx 134.42^\circ$, $C_2 \approx 15.58^\circ$, $c_2 \approx 3.76$.

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SSA Ambiguous Case Decision Chart