In ΔABC, prove that r₁ + r₂ + r₃ - r = 4R.
Answers
Answered by
10
Solution :
****************************************
We know that ,
r1 = 4RsinA/2cosB/2cosC/2
r2 = 4RcosA/2sinB/2cosC/2
r3 = 4RcosA/2cosB/2sinC/2
r = 4RsinA/2sinB/2sinC/2
************************************
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
••••
****************************************
We know that ,
r1 = 4RsinA/2cosB/2cosC/2
r2 = 4RcosA/2sinB/2cosC/2
r3 = 4RcosA/2cosB/2sinC/2
r = 4RsinA/2sinB/2sinC/2
************************************
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
••••
Answered by
4
Answer:
Step-by-step explanation:
We know:-,
r1 = 4RsinA/2cosB/2cosC/2
r2 = 4RcosA/2sinB/2cosC/2
r3 = 4RcosA/2cosB/2sinC/2
r = 4RsinA/2sinB/2sinC/2
now,
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
I HOPE IT'S HELP YOU DEAR,
THANKS
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