Properties & Solution of a Triangle

Notation & Basics

• In ΔABC: sides opposite to angles A, B, C are a, b, c respectively.
• Semi-perimeter: s = (a + b + c)/2
• Area: Δ

Sine Rule

• a/sinA = b/sinB = c/sinC
• Equivalently: sinA/a = sinB/b = sinC/c

Cosine Formulae (Law of Cosines)

• cosA = (b² + c² − a²)/(2bc)
• cosB = (c² + a² − b²)/(2ca)
• cosC = (a² + b² − c²)/(2ab)

Projection Formulae

• a = b cosC + c cosB
• b = c cosA + a cosC
• c = a cosB + b cosA

Half-Angle Formulae

• sin(A/2) = √[(s−b)(s−c)/(bc)]
• sin(B/2) = √[(s−c)(s−a)/(ca)]
• sin(C/2) = √[(s−a)(s−b)/(ab)]
• cos(A/2) = √[s(s−a)/(bc)]
• cos(B/2) = √[s(s−b)/(ca)]
• cos(C/2) = √[s(s−c)/(ab)]
• tan(A/2) = √[(s−b)(s−c)/(s(s−a))]
• tan(B/2) = √[(s−c)(s−a)/(s(s−b))]
• tan(C/2) = √[(s−a)(s−b)/(s(s−c))]

Area of a Triangle

• Δ = (1/2)bc sinA = (1/2)ca sinB = (1/2)ab sinC
• Δ = √[s(s−a)(s−b)(s−c)] (Heron’s formula)

Useful Area Relations

• sinA = 2Δ/(bc), sinB = 2Δ/(ca), sinC = 2Δ/(ab)
• sinA/a = sinB/b = sinC/c = 2Δ/(abc)

Napier’s Analogy

• tan((B−C)/2) = (b−c)/(b+c) · cot(A/2)
• tan((C−A)/2) = (c−a)/(c+a) · cot(B/2)
• tan((A−B)/2) = (a−b)/(a+b) · cot(C/2)

Circumcircle

• Circumradius: R = a/(2sinA) = b/(2sinB) = c/(2sinC) = abc/(4Δ)

Incircle

• Inradius: r = Δ/s
• r = (s−a)tan(A/2) = (s−b)tan(B/2) = (s−c)tan(C/2)
• r = 4R sin(A/2) sin(B/2) sin(C/2)

Excircles

• r₁ = Δ/(s−a), r₂ = Δ/(s−b), r₃ = Δ/(s−c)
• r₁ = s tan(A/2), r₂ = s tan(B/2), r₃ = s tan(C/2)
• r₁ = 4R sin(A/2) cos(B/2) cos(C/2) (cyclic for r₂, r₃)

Orthocentre

• Distances from vertices: OA = 2R cosA, OB = 2R cosB, OC = 2R cosC
• Distances to sides: OD = 2R cosB cosC, OE = 2R cosC cosA, OF = 2R cosA cosB
• Circumradius of pedal triangle: R/2
• Area of pedal triangle: 2Δ cosA cosB cosC

Important Results

• tan(A/2) − tan(B/2) = (s−c)/s and cot(A/2) − cot(B/2) = s/(s−c)
• tan(A/2) + tan(B/2) = c/s, cot(A/2) + cot(B/2) = (s−c)/Δ
• Cyclic identities with s, a, b, c are frequently tested in JEE Main.