MathIsimple

Triangle Solving

Master the Law of Sines, Law of Cosines, and triangle area formulas to solve any triangle problem with confidence.

Triangle Solver

Enter exactly 3 known values to solve the triangle. The calculator will determine the case and find all missing values.

Educational Mode: Sides must be integers, angles must be multiples of 15°. Results show radicals (√) and trigonometric values.

Enter 3 more values (currently 0/3)

Sides (integers only)

Angles (degrees) (multiples of 15°)

ABCcbaR

Core Concepts

Law of Sines
asin⁡A=bsin⁡B=csin⁡C=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R

The ratio of any side to the sine of its opposite angle is constant and equals the diameter of the circumscribed circle (circumradius R).

Derived Formulas

Find a side

a=b⋅sin⁡Asin⁡B=2R⋅sin⁡Aa = \frac{b \cdot \sin A}{\sin B} = 2R \cdot \sin A

Find an angle

sin⁡A=a⋅sin⁡Bb=a2R\sin A = \frac{a \cdot \sin B}{b} = \frac{a}{2R}

Circumradius

R=a2sin⁡A=abc4SR = \frac{a}{2\sin A} = \frac{abc}{4S}

When to Use:

  • AAS: Two angles and a non-included side
  • ASA: Two angles and the included side (find third angle first)
  • SSA: Two sides and a non-included angle (⚠️ ambiguous case)

The Ambiguous Case (SSA)

When given two sides (a, b) and an angle (A) opposite to one of them, there may be 0, 1, or 2 solutions:

If a < b·sin(A)→ No solution (no triangle exists)
If a = b·sin(A)→ Exactly one solution (right triangle)
If b·sin(A) < a < b→ Two solutions (ambiguous case)
If a ≥ b→ Exactly one solution

Example

Problem: In △ABC, a = 10, A = 30°, B = 45°. Find side b and the circumradius R.

Solution: b = a·sin(B)/sin(A) = 10×sin(45°)/sin(30°) = 10×0.7071/0.5 = 14.14. R = a/(2sin(A)) = 10/(2×0.5) = 10.

Law of Cosines

Generalizes the Pythagorean theorem to all triangles. When the included angle is 90°, it reduces to a² + b² = c².

Finding a Side (SAS)

Find side c (SAS)

c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C

Find side a

a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc\cos A

Find side b

b2=a2+c2−2accos⁡Bb^2 = a^2 + c^2 - 2ac\cos B

Finding an Angle (SSS)

Find angle C (SSS)

cos⁡C=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}

Find angle A

cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Find angle B

cos⁡B=a2+c2−b22ac\cos B = \frac{a^2 + c^2 - b^2}{2ac}

Determining Triangle Type

Compare c² with a² + b² (where c is the longest side):

c² < a² + b²→Acute triangle (all angles < 90°)
c² = a² + b²→Right triangle (one angle = 90°)
c² > a² + b²→Obtuse triangle (one angle > 90°)

When to Use:

  • SAS: Two sides and the included angle
  • SSS: All three sides known

Example

Problem: Triangle has sides a=5, b=7, c=8. Find angle C and determine the triangle type.

Solution: cos(C) = (25+49-64)/(2×5×7) = 10/70 = 1/7 ≈ 0.143, so C = arccos(0.143) ≈ 81.79°. Since 64 < 25+49=74, it's acute.

Triangle Area Formulas

Multiple methods to calculate triangle area depending on what information is available.

Base × Height

Base and perpendicular height known

S=12bhS = \frac{1}{2}bh

SAS Formula

Two sides and included angle

S=12absin⁡C=12bcsin⁡A=12acsin⁡BS = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B

Heron's Formula

All three sides known

S=s(s−a)(s−b)(s−c)S = \sqrt{s(s-a)(s-b)(s-c)}

Circumradius Formula

Three sides and circumradius

S=abc4RS = \frac{abc}{4R}

Inradius Formula

Inradius and semi-perimeter

S=rs=r⋅a+b+c2S = rs = r \cdot \frac{a+b+c}{2}

In Heron's formula, s is the semi-perimeter: s = (a + b + c)/2

Special Triangles

Equilateral Triangle

S=34a2S = \frac{\sqrt{3}}{4}a^2

Right Triangle

S=12×leg1×leg2S = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2

Isosceles Triangle

S=b44a2−b2S = \frac{b}{4}\sqrt{4a^2 - b^2}

Example

Problem: Find the area of a triangle with sides 13, 14, 15.

Solution: s = (13+14+15)/2 = 21. Area = √(21×8×7×6) = √7056 = 84 square units.

Quick Formula Reference

Law of Sines

asin⁡A=bsin⁡B=2R\frac{a}{\sin A} = \frac{b}{\sin B} = 2R

Law of Cosines (Side)

c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C

Law of Cosines (Angle)

cos⁡C=a2+b2−c22ab\cos C = \frac{a^2+b^2-c^2}{2ab}

Heron's Formula

S=s(s−a)(s−b)(s−c)S = \sqrt{s(s-a)(s-b)(s-c)}

Practice Quiz

Practice Quiz
10
Questions
0
Correct
0%
Accuracy
1
In △ABC, if a = 8, A = 40°, B = 60°, find side b using Law of Sines.
Not attempted
2
Using Law of Cosines, find angle C if a = 5, b = 6, c = 7.
Not attempted
3
What is the 'ambiguous case' in triangle solving?
Not attempted
4
Find the circumradius R if a = 10 and A = 30°.
Not attempted
5
Triangle with sides 8, 10, 13. Is it acute, right, or obtuse?
Not attempted
6
Find the area of △ABC with a = 6, b = 8, C = 30°.
Not attempted
7
What is the semi-perimeter of a triangle with sides 3, 4, 5?
Not attempted
8
Using Heron's formula, the area of a 3-4-5 triangle is:
Not attempted
9
If cos(C) = -0.5, what type of triangle is it?
Not attempted
10
Area of equilateral triangle with side 6 is:
Not attempted