How to solve a triangle from angle B, side a, and side c

Key takeaway: This looks like a standard Law of Sines problem, but it has the SSA flavor that can hide a second possibility. A careful angle check keeps you out of trouble.

We know:

• $B=16^\circ$
• $a=52$
• $c=114$

Because side $a$ is opposite $A$ and side $c$ is opposite $C$, start with the Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}.$ Use the known pair involving $B$ and $c$: $\frac{\sin C}{114}=\frac{\sin 16^\circ}{52}.$ So $\sin C=\frac{114\sin 16^\circ}{52}.$ Numerically, $\sin C\approx 0.6045,$ so $C\approx 37.2^\circ.$ The supplementary angle $142.8^\circ$ is also mathematically possible, so check both.

Case 1: $C\approx 37.2^\circ$

Then $A=180^\circ-16^\circ-37.2^\circ=126.8^\circ.$ Now find $b$: $\frac{b}{\sin 16^\circ}=\frac{52}{\sin 126.8^\circ},$ so $b\approx \frac{52\sin 16^\circ}{\sin 126.8^\circ}\approx 22.3.$

Case 2: $C\approx 142.8^\circ$

Then $A=180^\circ-16^\circ-142.8^\circ=21.2^\circ.$ Now $b\approx \frac{52\sin 16^\circ}{\sin 21.2^\circ}\approx 39.8.$

Result

This data actually allows two possible triangles:

• Triangle 1: $C\approx 37.2^\circ$, $A\approx 126.8^\circ$, $b\approx 22.3$
• Triangle 2: $C\approx 142.8^\circ$, $A\approx 21.2^\circ$, $b\approx 39.8$

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Pitfalls the pros know 👇 The biggest trap is assuming SSA always gives one triangle. It can give two. Another easy miss is mixing up which side is opposite which angle before applying the Law of Sines. If the labels slip, every later value comes out wrong.

What if the problem changes? If the problem had specified that triangle $ABC$ is acute, then only the smaller angle $C\approx 37.2^\circ$ would remain valid. If it had said $C$ is obtuse, then the supplementary angle would be the correct choice. Extra geometric information often decides the ambiguity.

Tags: Law of Sines, ambiguous case, SSA triangle