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Pre-Calculus Triangle trigonometry

Law of Sines (with ambiguous case)

20 practice questions 0 video lessons Theory + worked examples

Law of Sines

California Pre-Calculus • Standard G-SRT.11 • Triangle Trigonometry

Law of Sines is the opening topic of Triangle Trigonometry in the California Common Core State Standards. It is aligned to Standard G-SRT.11, which requires students to prove the Law of Sines and use it to solve problems.

The Law of Sines states \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) for any triangle, solving the AAS, ASA, and the ambiguous SSA cases.

California Pre-Calculus › Triangle Trigonometry › Law of Sines  —  Standard G-SRT.11

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Theory

The Law of Sines holds in any triangle, not just right triangles:

\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C},\]

where each side is labeled with the lowercase letter of its opposite angle. Use it when you know an angle and its opposite side, plus one more piece:

  • AAS or ASA — two angles and any side (always one triangle).
  • SSA — two sides and a non-included angle (the ambiguous case: zero, one, or two triangles).
The ambiguous case arises because a swinging side can reach the opposite line at two points; always check whether a second angle also works.
Law of Sines triangle An oblique triangle with vertices A, B, C and opposite sides a, b, c. a b c A B C
Each side pairs with its opposite angle: \(a\) with \(A\), etc.
The ambiguous case (SSA) With two sides and a non-included angle, a swinging side can reach the base at two points, giving two possible triangles. two ways to closeA
SSA: the side can close the triangle two ways, giving two solutions.

The Law of Sines, in both forms:

\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\quad\Longleftrightarrow\quad \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\]
a over sine A equals b over sine B equals c over sine C
Use the side-over-sine form to find a side, and the sine-over-side form to find an angle.

How to solve with the Law of Sines

  1. Match each side with its opposite angle.
  2. Set up a proportion using one complete pair and the unknown.
  3. Solve for the missing side or angle.
  4. For SSA, check the supplement of the angle you find — it may give a second valid triangle.
Example 1 — AAS: find a side
In a triangle, \(A=40^\circ\), \(B=75^\circ\), and side \(a=10\). Find side \(b\).
Solution

Use \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\) and solve for \(b\).

\(b\)\(=\)\(\dfrac{a\sin B}{\sin A}=\dfrac{10\sin 75^\circ}{\sin 40^\circ}\)
\(\approx\)\(15.0\)
side b is about 15.0
Example 2 — ASA: find the third angle first
With \(A=50^\circ\), \(C=60^\circ\), and included side \(b=8\), find side \(a\).
Solution

First \(B=180^\circ-50^\circ-60^\circ=70^\circ\), then apply the Law of Sines.

\(a\)\(=\)\(\dfrac{b\sin A}{\sin B}=\dfrac{8\sin 50^\circ}{\sin 70^\circ}\)
\(\approx\)\(6.5\)
side a is about 6.5
Example 3 — Find an angle
Given \(a=12\), \(b=9\), and \(A=55^\circ\), find angle \(B\).
Solution

Solve \(\dfrac{\sin B}{b}=\dfrac{\sin A}{a}\) for \(\sin B\).

\(\sin B\)\(=\)\(\dfrac{b\sin A}{a}=\dfrac{9\sin 55^\circ}{12}\)
\(\approx\)\(0.614\)
\(B\)\(\approx\)\(37.9^\circ\)
angle B is about 37.9 degrees
Example 4 — The ambiguous case
Why can \(a=6\), \(b=8\), \(A=35^\circ\) give two triangles?
Solution

This is SSA: the side opposite the known angle is shorter than the other known side, so the swinging side \(a\) can meet the base at two points.

\(\sin B\)\(=\)\(\dfrac{8\sin 35^\circ}{6}\approx 0.765\)
\(B\)\(\approx\)\(49.9^\circ\ \text{ or }\ 130.1^\circ\)

Both angles are valid (they keep the angle sum under \(180^\circ\)), giving two triangles.

angle B could be 49.9 or 130.1 degrees, two triangles

Common pitfalls

Check the ambiguous case. In SSA, the supplementary angle may also be valid — test whether the angle sum stays under \(180^\circ\).
Pair sides with opposite angles. The proportion only works when each side sits opposite its angle.
Find the third angle first in ASA. The given side may not be opposite a known angle until you do.

Frequently asked questions

What is the Law of Sines?

In any triangle, \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\); each side is proportional to the sine of its opposite angle.

When do you use the Law of Sines?

When you know an angle and its opposite side plus one more part: cases AAS, ASA, or SSA.

What is the ambiguous case?

The SSA situation, where the given data can produce zero, one, or two triangles. Always check whether a supplementary angle also fits.

How do you know if there are two triangles?

After finding an angle, test its supplement: if adding it to the known angle still leaves room under \(180^\circ\), a second triangle exists.