Side opposite to angle A is a,opp to B is b and opp to C is c

s=(a+b+c)/2    [semi perimeter]

cos A = (b²+c²-a²)/2bc

cos B = (c²+a²-b²⁾/2ac

cos C = (a²+b²-c²)/2ab

 

a=b cosC+c cosB

b=c cosA+a cosC

c=a cosB+b cosA

 

sin(A/2)=(s-b)(s-c)/bc

sin(B/2)=(s-a)(s-c)/ac

sin(C/2)=(s-a)(s-b)/ab

 

cos(A/2)=s(s-a)/bc

cos(B/2)=s(s-b)/ac

cos(C/2)=s(s-c)/ab

 

tan((B-C)/2)=(b-c)/(b+c) * cot(A/2)

tan((A-B)/2)=(a-b)/(a+b) * cot(B/2)

tan((C-A)/2)=(c-a)/(c+a) * cot(C/2)

 

Area of triangle = 1/2 ab sinC = 1/2 ac sinB = 1/2 bc sin A

                       =s(s-a)(s-b)(s-c)

 

a/sinA = b/sinB = c/sinC = 2R

R is radius of circumcircle of triangle ABC

 

R= abc/ 4 area of triangle

 

Inradius r

 

r= area of triangle/s

r= (s-a)tan(A/2)=(s-b)tan(B/2)=(s-c)tan(C/2)

 

r= asin(B/2)sin(C/2)/cos(A/2)

 = bsin(C/2)sin(A/2)/cos(B/2)

 = csin(A/2)sin(B/2)/cos(C/2)

 

r=4Rsin(A/2)sin(B/2)sin(C/2)

s=4Rcos(A/2)cos(B/2)cos(C/2)

 

Let r₁, r₂, r₃  be the exradii opp to A , B , C respectively

r₁=area/(s-a) = stan(A/2)

r₂=area/(s-b) = stan(B/2)

r₃=area/(s-c) = stan(C/2)

r₁=4Rsin(A/2)cos(B/2)cos(C/2)

r₂=4Rcos(A/2)sin(B/2)cos(C/2)

r₃=4Rcos(A/2)cos(B/2)sin(C/2)

 

H is the Orthocentre of the triangle(Altitudes intersect),then

 

AH=2RcosA

BH=2RcosB

CH=2RcosC

 

Let K, L,M be the pts where the altitudes respectively meet the opp side

 

HK=2RcosBcosC

HL=2RcosAcosC

HM=2RcosAcosB

 

G is the centroid ,O is the circumcentre,H is the orthocentre

HG/OG=2/1

 

Lenghts of bisectors of angles of triangle ABC=

2bc/(b+c) *cos(A/2), 2ca/(c+a)* cos(B/2) , 2ab/(a+b) *cos(C/2)

 

I is the Incentre of ABC

IA=r/sin(A/2)

IB=r/sin(B/2)

IC=r/sin(C/2)

I₁ , I₂ , I₃ be the centres of the Excircles opp to A B C resp

I₁A=r₁/sin(A/2)     I₂A=r₂/cos(A/2)    I₃A=r₃/cos(A/2)

I₁B=r₁/cos(B/2)    I₂B=r₂/sin(B/2)     I₃B=r₃/cos(B/2)

I₁C=r₁/cos(C/2)    I₂C=r₂/cos(C/2)    I₃C=r₃/sin(C/2)

 

II₁=a/cos(A/2)

II₂=b/cos(B/2)

II₃=c/cos(C/2)

 

II₁ * II₂ * II₃ = 16R²r

 

OH=R(1-8cosAcosBcosC)

OI=R(1-8sin(A/2)sin(B/2)sin(C/2))

HI=2r²-4R²cosAcosBcosC

 

If AD is the median of the triangle then

AB²+AC²=2(AD²+BD²)

 

In a cyclic quadrilateral ABCD

AB * CD +AD * BC= AC*BD

 

Following usual conventions

in a cyclic quadrilateral

cosB=(a²+b²-c²-d²)/2(ab+cd)

 

area = s(s-a)(s-b)(s-c)(s-d)

2s=a+b+c+d

 

Radius of circle circumscribing the quad

1/4 * (ab+cd)(ac+bd)(ad+bc)/(s-a)(s-b)(s-c)(s-d)

 

some basic formulae

sin18=((5)-1)/4

cos18=1/4(10+25))

 

If A+B+C=

sin2A+sin2B+sin2C=4sinAsinBsinC

sin2A+sin2B-sin2C=4cosAcosBsinC

cos2A+cos2B+cos2C= -1-4cosAcosBcosC

cos2A+cos2B-cos2C= 1-4sinAsinBsinC

sinA+sinB+sinC = 4 cos(A/2)cos(B/2)cos(C/2)

sinA+sinB-sinC= 4 sin(A/2)sin(B/2)cos(C/2)

cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sin(C/2)

cosA+cosB-cosC= -1+4cos(A/2)cos(B/2)sin(C/2)

tanA+tanB+tanC=tanAtanBtanC

cotAcotB+cotBcotC+cotCcotA=1

tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2)=1

cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)

tan2A+tan2B+tan2C=tan2Atan2Btan2C

tan(A₁+A₂+A₃+....A_n)= (S₁-S₃+S₅.......)/(1-S₂+S₄.....)

Where S₁,S₂,S₃,.... stand for sum taken 1 at time ,

2 at a time , 3 at a time and so on

 

Hope this will be useful

ALL THE BEST for everyone