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Question: In acute angled triangle ABC, r + r<sub>1</sub> = r<sub>2</sub> + r<sub>3</sub> and∠B \>\(\frac { \p...

In acute angled triangle ABC, r + r1 = r2 + r3 and∠B >π3\frac { \pi } { 3 }, then

A

b + 2c < 2a < 2b + 2c

B

b + 4c < 4a < 2b + 4c

C

b + 4c < 4a < 4b + 4c

D

b + 3c < 3a < 3b + 3c

Answer

b + 3c < 3a < 3b + 3c

Explanation

Solution

r – r2 = r3 – r1

−bs(s−b)=−a+c(s−a)(s−c)\frac { - b } { s ( s - b ) } = \frac { - a + c } { ( s - a ) ( s - c ) }

tan2(B/2) =

But Î(π6,π4)\left( \frac { \pi } { 6 } , \frac { \pi } { 4 } \right)

̃ tan2Î(13,1)\left( \frac { 1 } { 3 } , 1 \right)

̃ < 1

b < 3a – 3c < 3b

b + 3c < 3a < 3b + 3c