Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a triangle ABC, with known sides a=BC and b=AC, and known angle A, we wish to find angle B.

This is a typical application of the Sine Rule (Law of Sines).

In some circumstances, the sine rule gives an ambiguous result: with two possible solutions for angle B.

I am trying to find the simplest way of identifying whether or not the sine rule would give a unique solution.

Is it true to say that the sine rule will give a unique solution to this problem iff a > b?

share|cite|improve this question
use the law of cosines if you want $B$: $$b^2=a^2+c^2-2ac\cos B\quad\implies$$ $$B=\arccos\frac{a^2+c^2-b^2}{2ac}$$ For each $B\ne\frac{\pi}{2}=90^\circ$, there are two angles having $\sin B$; these depend on the length of $c$. So Law of Sines only gives unique $B$ when $\sin B=1$. – bgins Mar 29 '12 at 10:23
Thanks for your revision :) sin B = k may have two solutions but often one of those solutions will not allow a valid triangle: here, I don't consider that case to be a valid solution. – Ronald Mar 29 '12 at 10:29
This is, I think, much easier to approach by drawing diagrams of the geometry involved than by staring at the law of sines. – Chris Eagle Mar 29 '12 at 10:32
You're right. Then the question becomes - what are the necessary conditions on given sides a, b and angle A, to ensure uniqueness of the triangle to be drawn? – Ronald Mar 29 '12 at 10:36
up vote 3 down vote accepted

The question isn't really (or shouldn't be) about the sine rule, but about when two sides and an angle not formed by the two sides determine a unique triangle. You found almost the right criterion; in fact if $a=b$ the triangle is also uniquely determined (unless you allow degenerate triangles). The sine rule, by contrast, always allows two different angles at $B$, since the sine is symmetric with respect to reflection at $\pi/2$. If $a\ge b$, you can exclude the greater of the two because the sum with the angle at $A$ would exceed $\pi$, whereas for $a\lt b$ both of these angles correspond to triangles.

share|cite|improve this answer
The triangle is also uniquely determined if $a/b=\sin A$, and impossible if $a/b<\sin A$. – Chris Eagle Mar 29 '12 at 10:38
Thanks for your insight. – Ronald Mar 29 '12 at 10:39
@Chris: Ah, you're right; thanks. – joriki Mar 29 '12 at 10:53

This is also known as the ambiguous case (or SSA), and occurs whenever $a>b\,\sin A$, i.e., whenever $B\ne90^\circ$ and $b$ is not the hypotenuse (and $a$ the leg) of a right triangle.

enter image description here

Angle $B$ can be acute, $B_1=\arcsin\left(\frac{b}{a}\sin A\right)$, or obtuse, $B_2=\pi-B_1=180^\circ-B_1$, corresponding to which, side $c$ will be bigger ($c_1$) or smaller ($c_2$) than $b\,\cos A$, the common length from the right triangle.

share|cite|improve this answer

That is correct. (Or almost correct: there is no solution at all if $a < b \sin A$.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.