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Solving SSA Triangles — Definition, Formula & Examples

Solving SSA triangles means finding the unknown sides and angles of a triangle when you are given two sides and an angle opposite one of them. This arrangement is called the "ambiguous case" because the given information can produce zero, one, or two valid triangles.

Given a triangle with known sides aa and bb and a known angle AA opposite side aa, the SSA configuration requires solving sin⁡B=bsin⁡Aa\sin B = \frac{b \sin A}{a} for angle BB. Because the sine function yields values in both the first and second quadrants, the equation may have no solution (if bsin⁡Aa>1\frac{b \sin A}{a} > 1), exactly one solution (if BB must be acute or if the supplementary angle violates the triangle angle sum), or two distinct solutions.

Key Formula

sin⁡B=bsin⁡Aa\sin B = \frac{b \sin A}{a}
Where:
  • AA = The known angle (opposite side a)
  • aa = The side opposite angle A (known)
  • bb = The other known side (opposite unknown angle B)
  • BB = The unknown angle opposite side b

How It Works

Start by using the Law of Sines to find the sine of the unknown angle opposite the other known side. If sin⁡B>1\sin B > 1, no triangle exists. If sin⁡B≤1\sin B \leq 1, compute B=sin⁡−1 ⁣(bsin⁡Aa)B = \sin^{-1}\!\left(\frac{b \sin A}{a}\right). Then check whether the supplement B′=180°−BB' = 180° - B also forms a valid triangle (i.e., A+B′<180°A + B' < 180°). For each valid value of BB, find the third angle using C=180°−A−BC = 180° - A - B, then use the Law of Sines again to find the remaining side cc.

Worked Example

Problem: In triangle ABC, angle A = 30°, side a = 10, and side b = 16. Find all possible triangles.
Step 1: Apply the Law of Sines: Solve for sin B.
sin⁡B=bsin⁡Aa=16sin⁡30°10=16⋅0.510=0.8\sin B = \frac{b \sin A}{a} = \frac{16 \sin 30°}{10} = \frac{16 \cdot 0.5}{10} = 0.8
Step 2: Find angle B: Since sin B = 0.8 is between 0 and 1, at least one solution exists.
B1=sin⁡−1(0.8)≈53.13°B_1 = \sin^{-1}(0.8) \approx 53.13°
Step 3: Check the supplementary angle: The supplement is B₂ = 180° − 53.13° = 126.87°. Check: A + B₂ = 30° + 126.87° = 156.87° < 180°. This is valid, so there are two triangles.
B2=180°−53.13°=126.87°B_2 = 180° - 53.13° = 126.87°
Step 4: Solve Triangle 1 (B ≈ 53.13°): Find C and then side c using the Law of Sines.
C1=180°−30°−53.13°=96.87°,c1=asin⁡C1sin⁡A=10sin⁡96.87°sin⁡30°≈10⋅0.99250.5≈19.85C_1 = 180° - 30° - 53.13° = 96.87°, \quad c_1 = \frac{a \sin C_1}{\sin A} = \frac{10 \sin 96.87°}{\sin 30°} \approx \frac{10 \cdot 0.9925}{0.5} \approx 19.85
Step 5: Solve Triangle 2 (B ≈ 126.87°): Repeat for the second triangle.
C2=180°−30°−126.87°=23.13°,c2=10sin⁡23.13°sin⁡30°≈10⋅0.39280.5≈7.86C_2 = 180° - 30° - 126.87° = 23.13°, \quad c_2 = \frac{10 \sin 23.13°}{\sin 30°} \approx \frac{10 \cdot 0.3928}{0.5} \approx 7.86
Answer: Two triangles exist. Triangle 1: B ≈ 53.13°, C ≈ 96.87°, c ≈ 19.85. Triangle 2: B ≈ 126.87°, C ≈ 23.13°, c ≈ 7.86.

Another Example

Problem: In triangle ABC, angle A = 40°, side a = 5, and side b = 12. Find all possible triangles.
Step 1: Compute sin B: Apply the Law of Sines formula.
sin⁡B=12sin⁡40°5=12⋅0.64285≈1.543\sin B = \frac{12 \sin 40°}{5} = \frac{12 \cdot 0.6428}{5} \approx 1.543
Step 2: Interpret the result: Since sin B ≈ 1.543 > 1, no angle B exists. The sine of any angle cannot exceed 1.
Answer: No triangle exists with these measurements.

Why It Matters

The SSA ambiguous case appears throughout precalculus and trigonometry courses and is a staple on standardized exams like the AP Precalculus assessment. Surveyors and engineers routinely encounter SSA data when measuring distances to landmarks from a known baseline, making the ability to identify multiple solutions a practical skill, not just an academic exercise.

Common Mistakes

Mistake: Forgetting to check the supplementary angle B₂ = 180° − B₁
Correction: Always test whether 180° − B₁ also produces a valid triangle (A + B₂ < 180°). Skipping this step causes you to miss the second solution in the ambiguous case.
Mistake: Using the Law of Cosines formula for SSA and discarding a valid quadratic solution
Correction: If you set up the Law of Cosines as a quadratic in the unknown side, both positive roots may correspond to valid triangles. Be sure to evaluate every positive root, not just the first one.

Related Terms