Law of Sines: Complete Treatment

By the end of this lesson, you will be able to apply the Law of Sines to AAS, ASA, and SSA triangles, determine the number of solutions in the ambiguous case (0, 1, or 2 triangles), and find area using the sine formula.

The Law of Sines

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

Applies when: you have a matched angle-side pair (AAS or ASA). The SSA case is ambiguous.

The Ambiguous Case (SSA) — Complete Analysis

Given sides $a$, $b$ and angle $A$ (where $a$ is opposite $A$):

Case 1: $A \geq 90°$ (obtuse or right angle)

• $a > b$: exactly 1 triangle
• $a \leq b$: no triangle

Case 2: $A < 90°$ (acute angle)

Let $h = b\sin A$ (the altitude from vertex $C$ to side $AB$):

Condition	# Triangles
$a < h$	0
$a = h$	1 (right triangle)
$h < a < b$	2 (two distinct triangles)
$a \geq b$	1

Example 1 — AAS

$A = 35°$, $C = 80°$, $c = 15$. Find the remaining parts.

$B = 180 - 35 - 80 = 65°$

$\dfrac{a}{\sin 35°} = \dfrac{15}{\sin 80°}$

$a = \dfrac{15\sin 35°}{\sin 80°} \approx \dfrac{15(0.574)}{0.985} \approx 8.74$

$b = \dfrac{15\sin 65°}{\sin 80°} \approx \dfrac{15(0.906)}{0.985} \approx 13.80$

Example 2 — Ambiguous Case (Two Triangles)

$a = 8$, $b = 10$, $A = 40°$. Find all possible triangles.

$h = 10\sin 40° \approx 6.43$; since $h < a < b$ ($6.43 < 8 < 10$): two triangles.

$\sin B = \dfrac{b\sin A}{a} = \dfrac{10\sin 40°}{8} \approx 0.804$

$B_1 = \sin^{-1}(0.804) \approx 53.5°$ (acute); $B_2 = 180° - 53.5° = 126.5°$ (obtuse)

Triangle 1: $B \approx 53.5°$, $C = 86.5°$, $c = \dfrac{8\sin 86.5°}{\sin 40°} \approx 12.4$

Triangle 2: $B \approx 126.5°$, $C = 13.5°$, $c \approx 2.9$

Example 3 — No Solution

$a = 4$, $b = 10$, $A = 40°$.