Given A,a,b, first calculate s= bsin Aa. If s>1, no triangle exists. If 0<s 1, test both B_1=arcsin s and B_2=180 -B_1. Keep each distinct candidate only if A+B<180. When B_1=B_2=90, it is one triangle, not two. For acute A, write h=bsin A. There are two triangles precisely when h<a<b. Equality a=h gives one right triangle; a<h gives none. Example Let A=30, a=7, b=10. Then sin B= 57; B 45.6 or 134.4. Both leave a positive third angle, so both triangles are valid.

Given \(A,a,b\), first calculate \(s=\dfrac{b\sin A}{a}\). If \(s>1\), no triangle exists. If \(0<s\le1\), test both \(B_1=\arcsin s\) and \(B_2=180^\circ-B_1\).

Keep each distinct candidate only if \(A+B<180^\circ\). When \(B_1=B_2=90^\circ\), it is one triangle, not two.

For acute \(A\), write \(h=b\sin A\). There are two triangles precisely when \(h<a<b\). Equality \(a=h\) gives one right triangle; \(a<h\) gives none.

Let \(A=30^\circ\), \(a=7\), \(b=10\). Then \(\sin B=\dfrac57\); \(B\approx45.6^\circ\) or \(134.4^\circ\). Both leave a positive third angle, so both triangles are valid.

Trigonometry

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

The Ambiguous Sine-Rule Case

The Ambiguous Sine-Rule Case