How to solve a triangle when one angle and two sides are known

Expert intro: This is an SSA triangle, so the Law of Sines is the natural first move. It can feel straightforward, but you still want to check whether the data gives one triangle or more than one.

Detailed walkthrough

We are given:

• $A=16^\circ$
• $a=52$
• $b=114$

Because angle $A$ is opposite side $a$, use the Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}.$ Substitute the known values: $\frac{\sin 16^\circ}{52}=\frac{\sin B}{114}.$ Solve for $\sin B$: $\sin B=\frac{114\sin 16^\circ}{52}.$ Now compute: $\sin B\approx \frac{114(0.2756)}{52}\approx 0.6045.$ So $B\approx \sin^{-1}(0.6045)\approx 37.2^\circ.$ The supplementary angle is also possible: $180^\circ-37.2^\circ=142.8^\circ.$ But $16^\circ+142.8^\circ>180^\circ$, so that one is impossible.

Now find the third angle: $C=180^\circ-16^\circ-37.2^\circ=126.8^\circ.$

Use the Law of Sines again to find $c$: $\frac{c}{\sin C}=\frac{a}{\sin A},$ so $c=\frac{52\sin 126.8^\circ}{\sin 16^\circ}.$ This gives $c\approx 145.9.$

Final values

• $B\approx 37.2^\circ$
• $C\approx 126.8^\circ$
• $c\approx 145.9$

💡 Pitfall guide

With SSA data, students often stop after the first inverse sine and forget to test the supplementary angle. Here, the triangle sum rules it out, but in other problems the second triangle can be valid. Also, keep side labels matched to opposite angles: $a\leftrightarrow A$, $b\leftrightarrow B$, $c\leftrightarrow C$.

🔄 Real-world variant

If $b$ had been much smaller, the sine ratio could produce no triangle at all. If $b$ had been different but still valid, you might get two possible triangles from the same setup. The quickest way to check is: after finding one angle with inverse sine, test the supplement and see whether the remaining angles still add to less than $180^\circ$.

🔍 Related terms

Law of Sines, SSA triangle, supplementary angle