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Solving Triangles

To solve a triangle means to find all its unknown sides and angles given sufficient information. The Laws of Cosines and Sines allow us to solve any triangle.

We use the standard notation: sides aa, bb, cc opposite to angles α\alpha, β\beta, γ\gamma respectively.

Interactive Solver

Select a case, set the known values, and see the triangle solved in real time. For SSA, watch how two valid triangles can appear:

b = 3.0
c = 4.0
α (included) = 60.0°
a = 3.61β = 46.1°γ = 73.9°

Case 1: One Side and Two Angles

Given aa, β\beta, γ\gamma: find α=180°−(β+γ)\alpha = 180° - (\beta + \gamma), then use the Law of Sines.

✎Example — Side and Two Angles

Solve triangle with a=12a = 12 cm, β=36°\beta = 36°, γ=119°\gamma = 119°.

Solution. α=180°−(36°+119°)=25°\alpha = 180° - (36° + 119°) = 25°.

By the Law of Sines: b=asin⁡βsin⁡α=12sin⁡36°sin⁡25°≈12⋅0.590.42≈16.9 cmb = \frac{a\sin\beta}{\sin\alpha} = \frac{12\sin 36°}{\sin 25°} \approx \frac{12 \cdot 0.59}{0.42} \approx 16.9 \text{ cm} c=asin⁡γsin⁡α=12sin⁡119°sin⁡25°=12sin⁡61°sin⁡25°≈12⋅0.870.42≈24.9 cmc = \frac{a\sin\gamma}{\sin\alpha} = \frac{12\sin 119°}{\sin 25°} = \frac{12\sin 61°}{\sin 25°} \approx \frac{12 \cdot 0.87}{0.42} \approx 24.9 \text{ cm}

Answer: b≈16.9b \approx 16.9 cm, c≈24.9c \approx 24.9 cm, α=25°\alpha = 25°. ◀\blacktriangleleft

Case 2: Two Sides and the Included Angle (SAS)

Given bb, cc, α\alpha: find aa by the Law of Cosines, then find the remaining angles.

✎Example — Two Sides and Included Angle

Solve: a=14a = 14 cm, b=8b = 8 cm, γ=38°\gamma = 38°.

Solution. By Law of Cosines: c2=a2+b2−2abcos⁡γ=196+64−2⋅14⋅8cos⁡38°≈260−224⋅0.79≈83c^2 = a^2 + b^2 - 2ab\cos\gamma = 196 + 64 - 2\cdot14\cdot8\cos38° \approx 260 - 224\cdot0.79 \approx 83 c≈9.1 cmc \approx 9.1 \text{ cm}

Then cos⁡α=b2+c2−a22bc≈64+83−1962⋅8⋅9.1≈−0.34\cos\alpha = \frac{b^2+c^2-a^2}{2bc} \approx \frac{64+83-196}{2\cdot8\cdot9.1} \approx -0.34, so α≈110°\alpha \approx 110°, β≈32°\beta \approx 32°. ◀\blacktriangleleft

Case 3: Three Sides (SSS)

Given aa, bb, cc: find angles using the Law of Cosines.

✎Example — Three Sides Given

Solve: a=7a = 7 cm, b=2b = 2 cm, c=8c = 8 cm.

Solution. cos⁡α=b2+c2−a22bc=4+64−4932≈0.59  ⟹  α≈54°\cos\alpha = \frac{b^2+c^2-a^2}{2bc} = \frac{4+64-49}{32} \approx 0.59 \implies \alpha \approx 54°

By Law of Sines: sin⁡β=bsin⁡αa≈2⋅0.817≈0.23  ⟹  β≈13°\sin\beta = \frac{b\sin\alpha}{a} \approx \frac{2\cdot0.81}{7} \approx 0.23 \implies \beta \approx 13°.

Then γ=180°−(54°+13°)=113°\gamma = 180° - (54° + 13°) = 113°. ◀\blacktriangleleft

Case 4: Two Sides and a Non-Included Angle (SSA — Ambiguous Case)

Given aa, bb, α\alpha (angle opposite to aa): this may yield 0, 1, or 2 solutions.

✎Example — The Ambiguous Case

Solve: a=6a = 6 cm, b=5b = 5 cm, β=50°\beta = 50°.

Solution. By Law of Sines: sin⁡α=asin⁡βb=6⋅0.775≈0.92\sin\alpha = \frac{a\sin\beta}{b} = \frac{6\cdot0.77}{5} \approx 0.92.

Possible values: α≈67°\alpha \approx 67° or α≈113°\alpha \approx 113°. Both are valid since α>β\alpha > \beta in both cases.

Case A: α≈67°\alpha \approx 67°, γ≈63°\gamma \approx 63°, c≈5.8c \approx 5.8 cm. Case B: α≈113°\alpha \approx 113°, γ≈17°\gamma \approx 17°, c≈1.9c \approx 1.9 cm. ◀\blacktriangleleft

Summary

CaseGivenMethodSolutions
1: Side and Two Anglesaa, β\beta, γ\gammaα=180°−(β+γ)\alpha = 180° - (\beta+\gamma), then Law of SinesExactly 1
2: Two Sides and Included Anglebb, cc, α\alphaLaw of Cosines for aa, then Law of SinesExactly 1
3: Three Sidesaa, bb, ccLaw of Cosines for each angleExactly 1
4: Two Sides and Non-Included Angleaa, bb, α\alphaLaw of Sines for sin⁡β\sin\beta, check validity0, 1, or 2