IIT JEE , AIEEE Forum - Mathematics , Trignometry

vandya

In a triangle ABC prove
(ab-r1r2)/r3=r
where r1,r2,r3 are the exradii and r is the inradius

prash_shan_jpr

First make it

ab+r1r2 = r3 r

ab = r3 r - r1 r2

ok

now put r1 = delta/(s-a)

r2 = delta/(s-b)

r3 = delta/(s-c)

r = delta/s

ok

now solve it

u will find that RHS = LHS

LHS = ab

in 3 steps u can get the ans.

any more doubtsss..?

sboosy

to prove :

ab = rr₃ + r₁r₂

LHS= 2RsinA *2RsinB=4R²sinAsinB

RHS:

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

r₃=4Rcos(A/2)cos(B/2)sin(C/2)

r₂=4Rcos(A/2)sin(B/2)cos(C/2)

r₁=4Rsin(A/2)cos(B/2)cos(C/2)

so rr₃= 4R²sin²(C/2) sinAsinB

r₁r₂=4R²cos²(C/2) sinAsinB

adding both

4R²sinAsinB [ sin²(C/2)+cos²(C/2)]

= 4R²sinAsinB

LHS = RHS

hence proved