Question

in triangle a,b,c prove that r1+r2+r3-r=4R​

Answer 1

Answer:

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Step-by-step explanation:

We know that ,

r1 = 4RsinA/2cosB/2cosC/2

r2 = 4RcosA/2sinB/2cosC/2

r3 = 4RcosA/2cosB/2sinC/2

r = 4RsinA/2sinB/2sinC/2

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LHS = r1 + r2 + r3 - r

= 4RsinA/2cosB/2cosC/2

+ 4RcosA/2sinB/2cosC/2

+ 4RcosA/2cosB/2sinC/2

- 4RsinA/2sinB/2sinC/2

=4RcosC/2[sinA/2cosB/2+cosA/2sinB/2]

+4RsinC/2[cosA/2cosB/2-sinA/2sinB/2]

= 4RcosC/2sin[(A+B)/2]

+ 4RsinC/2cos[(A+B)/2]

= 4R{cosC/2sin[(A+B)/2]+cos[(A+B)/2sinC/2}

= 4R sin[ ( A+B+C)/2 ]

= 4R sin 90°

= 4R × 1

= 4R

= RHS