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If F is the mid point of AB, then AB/2=BF=AF, AB=2BF=2AF
Ab should be twice of BF and AF

can u plz tell me hoe the ans is sum of roots taken 2 at a time?

the resultant of two vectors A and B is root(A^2+B^2 +2ABcos@)
where @ is the angle between them
now, Let AB=1
BC=2
angle between them is 120
resultant is root(1+4+4cos120)=root(3)
and the angle which this resultant makes with AB=arctan2sin120/1+2cos120=arctaninfinity =pi/2
and total angle with AB=270
the angle which AB makes with AC=120
angle between the resultant and AC=60+90=150
resulatnt =root(3+9+6root3cos150)
=root(12+6root3(-root3/2))

root(12-9)=root3
hence the ans is C

how is AB=BF/2=AF/2
AB/2=BF=AF

can u plz post the soln

is the triangle equilateral traiangle or not

can u plz tell me how the other angles are 120-A/2 and 60-A/2.
and also this step
4*(Cot(120-A/2)+Cot(60-A/2))=6

is A the vertex angle?

if a vector subtends some angle with the x axis, some other angles with y and z axes, then it is impossible to determine the other two if we know only one
so, for simplicity, i have assumed that the vector A subtends the angle a with positive x axis, vector (and 90-a with positive y axis and it lies only in xy plane)
vector B makes an angle of b with positive y xais and it lies completly in yz plane and the vector C too lies completly in the yz plane making an angle of c with positive z xais
as the total angle between the y and z axes is 90, let us assume that the angle between the vectors C and B to be 90-(b+c)
the y component of vector a is Asina and the x component is Acosa
now taking the component of vector Asina along the vector B, we get, Asinacosb
now finding the resultant between Asinacosb+B and C, we get
root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))
and the angle this resultant makes with positive y axis is arctan({[Ccos(b+c)]/[Asinacosb+B+Csin(b+c)] +b})
the angle which this resultant makes with positive z axis is
arc tan((Asinacosb+B)cos(b+c)/[p+(Asinacosb+B)sin(b+c)]n +c)

now let us find the vector sum of this resultant with Acosa (which is along x axis)
root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c)+A^2cos^2a
angle which this new resulatnt makes with x axis is
arctan{root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))}/[Acosa]
the cos of angle which this new resulatnt makes with positive y axis is cos(arctanAcosa/root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))cos(arctan(Pcos(b+c)/[(Asinacosb+B)+Psin(b+c)] +b)
the cos of angle which this new resulatnt makes with positive z axis is
cos(arctanAcosa/root((Asinacosb+B)^2+C^2+2C(Asinacosb+B)sin(b+c))cos(arctan(Asinacosb+B)cos(b+c)/[P+(Asinacosb+B)sin(b+c)] + c)
we know that cos^2@ +cos^2#+cos^2$=1
but i ma not understanding how i can get the value of sina+sinb+sinc from this realtion
i have written the eqaution for cos^2@ +cos^2#+cos^2$=1
but it took two pages to write it down
i dont know whether i can solve or not
if possible i will post the soln tomorrow
if u know the soln , plz post it

am i correct?

sorry, the values of y are coming out to be -0.44,-0.2481(along with two imaginary roots)
but p and m cannot be negative
also, if we take the question into consideration,
2(bc^2+ca^2+ab^2)=b^2c+c^2a+a^2b+3abc
rearranging,
c(2a^2-b^2)+a(2b^2-c^2)=-2bc^2+a^2b+3abc
putting the value of b^2 as ac
[c(2a^2-ac)+a(2ac-c^2)]/[-2c^2+a^2+3ac]=b
[4a^2c-2ac^2]/[-2c^2+a^2+3ac]=b
2ac(2a-c)/[-2c^2+a^2+3ac]=b
substituting the values of a, b, c as a,ai/2,ai^2/4
we finally get
i^4-2i^3-6i^2+16i-8=0
but this is different fron the earler equation y^4-2y^3-6y^2+16y+8=0
where i is one of the soln of y
only the sign of last term is different
cant understand y
salute assured 4 the best reply (with proper reasoning0
now if we substitute i=2 in i^4-2i^3-6i^2+16i-8=0
it satisfies the equation
and also i=2 makes a=b=c
so acording to the condition the triangle is equilateral

i actually thought of a different method

2(bc^2+ca^2+ab^2)=b^2c+c^2a+a^2b+3abc

2abc(b/c+c/a+a/b)=abc(b/a+a/c+c/b+3)

put c/b=p

2(1/p+pb/a+a/b)=(b/a+a/pb+p+3)

put b/a=m

2(1/p+pm+1/m-1)=(m+1/mp+p+1)

2((m+p+p^2m^2-pm)=m^2p+1+mp^2+mp

m(2-mp)+p(2-mp)+mp((2mp-1)-(2mp+1)

put mp=r

(m+p)(2-r)+r(2r-1)-(2r+1)

put m+p=y

2r^2-r(3+y)+2y+1

solving for r

[(3+y)+/-root(1+y^2-10y)]/4

considering only the positive value(just for simplicity)

[(3+y)+root(1+y^2-10y)]/4=r............................................j

we know that m+p=y

mp=r

(m-p)^2=(m+p)^2-mp

(m-p)=root(y^2-4r)

m=[root(y^2-4r)+y]/2..........................................................f

p=2r/[root(y^2-4r)+y] (as mp=r)...........................................1

but from the equation p=[y-root(y^2-4r)]/2..................................g

from f

2m-y=[root(y^2-4r)

(2m-y)^2=(y^2-4r)

from g

2p-y=-root(y^2-4r)

(2p-y)^2=(y^2-4r)

from these two results

p=m is one of the soln

equating 1 and f

2r/[root(y^2-4r)+y] =[root(y^2-4r)+y]/2

4r=y^2-4r+y^2+2y(root(y^2-4r)

4r-y^2=y(root(y^2-4r)

squaring on both sides

16r^2+y^4-8ry^2=y^2(y^2-4r)

16r^2+y^4-8ry^2=y^4-4ry^2

4r=y^2

substituting the value or r from j

(3+y)+root(1+y^2-10y)=y^2

3+y-y^2=-root(1+y^2-10y)

9+y^4+6y-2y^3-6y^2=1-10y

y^4-2y^3-6y^2+16y+8=0

let t,y,u,i be the solutions of y(like i am not getting nice integers after trying with calculator)

we know that y=p+m

and that p=m

p=m=y/2

p=c/b and m=b/a

as p=m , b^2=ac

also mp=r=c/a

p^2=m^2=c/a

let p=m=i/2(putting one of the soln of y)

i^2/4=c/a

4c=ai^2

we know that b^2=ac

b^2=aai^2/4

b=ai/2(considering only the positive value)

so the sides are related as a,b,c=a,ai/2,ai^2/4, it can be an equilateral triangle if i=2

but putting the value of i in y^4-2y^3-6y^2+16y+8=0, it is not coming out to be 0

for this to be a right angled triangle

a^2i^2/4=a^2i^4/16+a^2

putting i^2/4=h

h=h^2+1

solving, we get imaginary roots, so this is not a rt angled triangle too

it is not coming ut to be 0 , so it is not a rt angled triangle

for it to be isosceles, any of these three conditions should be satisfied a=ai/2,ai/2=ai^2/4,a=ai^2/4

let us take the first condition a=ai/2

i=2 is not the soln

so , it is rooted out

ai/2=ai^2/4

it is also rooted out

so, we are left with only

a=ai^2/4

i=+/-2

substituting y=-2 in

y^4-2y^3-6y^2+16y+8=0

it does not satisfy the equation

hence , it is not an isosceles nor equilateral triangle nor a rt triangle

the value of y is nearly equal to 0.45

i will post the soln tomorrow

if the lengths of sides of the triangle ABC satisfy

2(bc2+ca2+ab2)=b2c+c2a+a2b+3abc then the triangle is ?

2(bc^2+ca^2+ab^2)==b^2c+c^2a+a^2b+3abc

ab(2b-a)+ca(2a-c)+bc(2c-b)-abc-abc-abc=0

ab(2b-a-c)+ca(2a-c-b)+bc(2c-b-a)=0

let b-a=p

ab(b+p-c)+ca(a-c-p)+bc(2c-2b+p)=0

a(b^2+pb-bc+ac-c^2-cp)=-bc(2c-2b+p)

a(-2cp+pb+b^2-c^2)=-bc(2c-2b+p)

-2cpa+pba+ab^2-ac^2+2bc^2-2b^2c+pbc=0

pc(-2a+b)+pba+ab^2-ac^2+2bc^2-2b^2c=0

p^2c-apc+pba+ab^2-ac^2+2bc^2-2b^2c=0

a(b+c)(b-c)+2bc(c-b)+p^2c+ap(b-c)

(b-c)[a(b+c)-2bc+ap)+p^2c=0

(b-c)(ab+(b-p)c-2bc+ap)+p^2c=0

(b-c)(b+p)(a-c)+p^2c=0

i am unable to think further

i will post the remaining soln 2morow

even i want the soln,
can anyone plz ans
rates assured

i dont know the ans, can u plz post the soln