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=

2r
2-4R
2cosAcosBcosC
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+2

5))
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+....An)= (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