Question

In ΔABC, prove that r + r₁ + r₂ - r₃ = 4R cos C.

Answer 1

Solution :

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We know that ,

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

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

iii ) r2= 4RcosA/2sinB/2sinC/2

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

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LHS = r + r1 + r2 - r3

= 4RsinA/2sinB/2sinC/2

+ 4RsinA/2cosB/2cosC/2

+ 4RcosA/2sinB/2scosC/2

- 4RcosA/2cosB/2sinC/2

= 4RsinA/2[cosB/2cosC/2+sinB/2sinC/2]

+4RcosA/2[sinB/2cosC/2-cosB/2cosC/2]

= 4RsinA/2cos[(B-C)/2]

+ 4RcosA/2sin[(B-C)/2]

= 4Rcos[(B+C)/2] cos [(B-C)/2]

+ 4Rsin[(B+C)/2]sin[(B-C)/2]

= 4R{ cos[(B+C)/2cos[(B-C)/2

+ sin[(B+C)/2]sin[(B-C)/2]

= 4R cos [ (B+C)/2 - ( B-C)/2 ]

= 4R cos [ ( B+C- B + C )/2 ]

= 4R cos ( 2C/2 )

= 4R cos C/2

= RHS

••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

We know that ,

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

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

r2= 4RcosA/2sinB/2sinC/2

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

now, r + r1 + r2 - r3

= 4RsinA/2sinB/2sinC/2

+ 4RsinA/2cosB/2cosC/2

+ 4RcosA/2sinB/2scosC/2

- 4RcosA/2cosB/2sinC/2

=> 4RsinA/2[cosB/2cosC/2+sinB/2sinC/2]

+4RcosA/2[sinB/2cosC/2-cosB/2cosC/2]

=> 4RsinA/2cos[(B-C)/2]

+ 4RcosA/2sin[(B-C)/2]

=> 4Rcos[(B+C)/2] cos [(B-C)/2]

+ 4Rsin[(B+C)/2]sin[(B-C)/2]

=> 4R{cos[(B+C)/2cos[(B-C)/2

+ sin[(B+C)/2]sin[(B-C)/2]

=> 4R cos [ (B+C)/2 - ( B-C)/2 ]

=> 4R cos [ ( B+C- B + C )/2 ]

=> 4R cos ( 2C/2 )

=> 4R cos C/2

I HOPE IT'S HELP YOU DEAR,

THANKS