# Law of Sines will give a unique solution iff a > b?

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?

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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

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.

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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.

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.

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

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