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Community shelf Community shelf -> Solutions to Triangles -> Go to message
This Post 54 points    (10 Olaaa!! Perrrfect answer.   in 12 votes )   [?]
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+....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
Catalogs Discussion Forums -> Integral Calculus -> solve sec^3 and tan^3...required for theory -> Go to message
This Post 2 points    (Olaaa!! Perrrfect answer.   in 1 votes )   [?]
sec3x dx =  secx sec2x dx
tan x =t
sec2x dx = dt
sec x = 1+t2
use the formula for
 1+t2 dt = (t1+t2 )/2 + 1/2(log (t+1+t2) + c
Catalogs Discussion Forums -> Trignometry -> Mathematics buk -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
ARIHANT
though u need to b careful 'at times' because it could be wrong
but otherwise it is one of the best
specifically if u want
for calculus: shanthinarayanan
then books by I A Maron
for Trigonometry S L Loney
and ofcourse its not possibe to list them all
so if u want any other topic in particular then ask
Catalogs Discussion Forums -> Integral Calculus -> o to infinity dx/(a*a + x*x) to the power 7. -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
putting x=atan(b)
limit changes from 0 to pi/2
dx=asecb*secb db
substituing
weget
integral of o to pi/2
cos^12b(cos power 12 b) db/ a^13
so using reduction formula
ans:
1/a^13 * [11*9*7*5*3*1] / [12*10*8*6*4*2] * pi/2
Catalogs Discussion Forums -> Algebra -> geom(inmo) -> Go to message
This Post 17 points    (Olaaa!! Perrrfect answer.   in 4 votes )   [?]
let the usual symbols stand for what they are
that is sides of the triangle are a,b,c
now rt angled at A means a*a=b*b+c*c .....1
the two mentioned lines meet at I the symbol usually used
now area of triangle IBC
drop a perpendicular IK on BC(actually the inradius)
now required area = area of IKB+area of IKC
IK=r,KB=rcot(B/2),KC=rcot(C/2)
total area = 1/2 * r * r *(cotB/2 +cotC/2)
using r= area of abc / semi perimeter we get it equal to
b*b*c*c/(a+b+c)*(a+b+c)
cotB/2=(a+c)/b
cotC/2=(a+b)/c
thus we get area of IBC
now area of bcde=
area of ABC - area of triangle ADE
area ABC = 1/2 *b* c
area of ADE =
1/2 * AD *AE
using angle bisector theorem
a/b=BE/EA
thus EA=bc/(a+b)
similarly AD=bc/(a+c)
now substitue
for simplification use the following
(a+b+c)*(a+b+c)=2[b*b+c*c+ab+bc+ca]
bcos of 1
thus the result
....nice question
Catalogs Discussion Forums -> Algebra -> PROBABILITY -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
Priyesh is spot on
p(A union B)=p(A)+1-p(not B)-p(A)[1-p(not B))
0.85=0.35+P(a)(0.65)
p(A)=10/13
Catalogs Discussion Forums -> Integral Calculus -> Integrals Challenge : rates for best answer -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
Bhattji at ur very best superb
Catalogs Discussion Forums -> Integral Calculus -> DEFINITE integration 8 -> Go to message
This Post 25 points    (Olaaa!! Perrrfect answer.   in 5 votes )   [?]
integral
[t+1]cubed
=([t]+1)whole cubed using the fact that
[n+k]=[n]+k where k is a constant
so integral 0to x can be written as o to [x] and
[x] to x
first integral we get 1cube +2 cube+....[x] cubed
using formula it is first part of answer
next one
[x] to x the value of [x]+1 is constant
so that multipled with x-[x]={x}
so thats the answer
Catalogs Discussion Forums -> Coaching Institutes & Course Material -> FIITJEE AITS-Rubbish, garbage -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
Brilliant i completely agree
Catalogs Discussion Forums -> Trignometry -> please -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
What do u mean by trignometry in physics and chemistry?
 
Catalogs Discussion Forums -> Coaching Institutes & Course Material -> Percentile Rank Prediction(FIITJEE) -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
Never see the marks
Ultimately it is how u write the final exam that matters
so keep reading thats all
Catalogs Discussion Forums -> Coaching Institutes & Course Material -> FIITJEE AITS-Rubbish, garbage -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
Whoever says some answers are wrong i completely agree with him
Whoever says questions are at a nice level ideal to crack jee with calm
i agree with him
The simple point is not to fuss too much solutions.Instead try to get the right
answer for those questions u feel answer is given wrong by posting in probably this same site or elsewhere and continue reading for the forthcoming exams.
Count those marks in ur favour if u think u r right
So what is so difficult abt this? 
Catalogs Discussion Forums -> Differential Calculus -> Limit -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
am i right or not?
Catalogs Discussion Forums -> Differential Calculus -> Limit -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
I think answer is 0.i applied lhospital rule
so just differentiating the numerator i get ans=0
pls correct me if wrong
Catalogs Discussion Forums -> Algebra -> probability problem -> Go to message
This Post 0 points    (Olaaa!! Perrrfect answer.   in 0 votes )   [?]
BIpin sir 5 and 10 not included
 
 
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