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 15Β°)

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=bβ‹…sin⁑Asin⁑B=2Rβ‹…sin⁑Aa = \frac{b \cdot \sin A}{\sin B} = 2R \cdot \sin A

Find an angle

sin⁑A=aβ‹…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 < bΒ·sin(A)β†’ No solution (no triangle exists)
If a = bΒ·sin(A)β†’ Exactly one solution (right triangle)
If b·sin(A) < a < b→ Two solutions (ambiguous case)
If a β‰₯ bβ†’ Exactly one solution

Example

Problem: In β–³ABC, a = 10, A = 30Β°, B = 45Β°. Find side b and the circumradius R.

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

Law of Cosines

Generalizes the Pythagorean theorem to all triangles. When the included angle is 90Β°, it reduces to aΒ² + bΒ² = cΒ².

Finding a Side (SAS)

Find side c (SAS)

c2=a2+b2βˆ’2abcos⁑Cc^2 = a^2 + b^2 - 2ab\cos C

Find side a

a2=b2+c2βˆ’2bccos⁑Aa^2 = b^2 + c^2 - 2bc\cos A

Find side b

b2=a2+c2βˆ’2accos⁑Bb^2 = a^2 + c^2 - 2ac\cos B

Finding an Angle (SSS)

Find angle C (SSS)

cos⁑C=a2+b2βˆ’c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab}

Find angle A

cos⁑A=b2+c2βˆ’a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Find angle B

cos⁑B=a2+c2βˆ’b22ac\cos B = \frac{a^2 + c^2 - b^2}{2ac}

Determining Triangle Type

Compare cΒ² with aΒ² + bΒ² (where c is the longest side):

cΒ² < aΒ² + bΒ²β†’Acute triangle (all angles < 90Β°)
cΒ² = aΒ² + bΒ²β†’Right triangle (one angle = 90Β°)
cΒ² > aΒ² + bΒ²β†’Obtuse triangle (one angle > 90Β°)

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)/(2Γ—5Γ—7) = 10/70 = 1/7 β‰ˆ 0.143, so C = arccos(0.143) β‰ˆ 81.79Β°. Since 64 < 25+49=74, it's acute.

Triangle Area Formulas

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

Base Γ— 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(sβˆ’a)(sβˆ’b)(sβˆ’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=rβ‹…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=12Γ—leg1Γ—leg2S = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2

Isosceles Triangle

S=b44a2βˆ’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 = √(21Γ—8Γ—7Γ—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+b2βˆ’2abcos⁑Cc^2 = a^2 + b^2 - 2ab\cos C

Law of Cosines (Angle)

cos⁑C=a2+b2βˆ’c22ab\cos C = \frac{a^2+b^2-c^2}{2ab}

Heron's Formula

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

Practice Quiz

Practice Quiz
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