Question

In a triangle ABC prove that r+r1+r2-r3=4Rcosc

Answer 1

Explanation:

In the given question we have to use property of triangles

r =

s

Δ

, r1 =∆ / s − a/, r2 = ∆/s − b, r3 = ∆/ s − c

r+r1+r2-r3=4Rcosc

LHS can be expressed as

=∆/s + ∆/s-a + ∆/s-b - ∆/s-c

=∆(1/s+ 1/s-a + 1/s-b - 1/s-c)

= ∆(1/s - 1/s-c )+∆/ (1/s-a -l+ 1/s-b)

=∆(1/s-b+ 1/s-a + 1/s - 1/s-c)

=∆[c/(s-b)(s-a )+ c/s*(s-c)]

2s-a-b=c as a+b+c is semi perimeter or a+b+c/2

= c∆[{2s²-s(a+b+c)+ab}/∆²]

=abc/∆

=4R

We also know 4R = 2ccosecc= 2cosc

As 2R= c/sinc

Also R= abc/4∆

So,4R= abc/∆

Hence LHS= 2cosc