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Question

In Δ ABC the sides opposite to angles A,B,C are denoted by a,b,c respectively.
The lengths of the tangents from A,B and C to the incircle are in A.P., then?

A
r1,r2,r3 are in H . P.
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B
r1,r2,r3 are in A . P.
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C
a,b,c are in A . P.
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D
cosA=4c3b2c
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Solution

The correct options are
A r1,r2,r3 are in H . P.
C a,b,c are in A . P.
D cosA=4c3b2c
such that Zy=(yx)or(yz)=(xy)(1)s=a+b+c2=2(x+y+z)2=s=y+2y=3y(ii)a=y+z,b=x+z,c=x+y(iii)ba(xy),cb=(yz)
From (1) ba=cb2b=a+c
a,b,c, are in A.P.
r1,r2,r3 are the ex centre of triangle
we know r1=s(sc)(sb),r2=S(sa)(sc),r3=s(sa)(sb)
from (1),(11),(111)
r1=(3y).[3y(y+X)][3y(x+y)]=(3y)[2yX].[y]r2=(3y)[3y(y+z)][3y(x+y)]=(3y)(2yz)(2yX)r3=(3y)[3y(y+z)][3y(x+y)]=(3y)(2yz)(y)1r1+1r3=3y(y)[1(24X)+1(24Z)]=(3y2)[4y(x+x)(24X)(24z)]=(4y2y)(3y)(y).(24X)(24z)=2y(3y)(y)(24X)(24z)=zr3r1,r2,r3cosA=b2+c2a22bc=(X+Z)2+(x+4)2(y+z)22(x+z)(c)=(x+z)2+(xz)(x+z)+2y)2.(x+z)C=(x+z)2+(xz)(x+z)+(X+Z)2.(x+z)C=(X+Z)(x+z)+(XZ)2.(X+Z)C=XZ2C==4x+4x+2Z3(X+Z)2C=4X+2(2y)3b2C=4(X+4)3b2C=4C3b2C=4(X+4)3b2C=4C3b2C

906443_142435_ans_fdc194576ef44c5bbe92d45622141195.png

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