Question

In a triangle ABC r=R1+r3-r2 And angle B>pi/3 then the range of s/b

Answer 1

b + 3c < 3a

Step-by-step explanation:

The question is incomplete. The full question is:

In an acute angled triangle ABC,r+r1=r2+r3and∠B>π3, then b+2c<2a<2b+2c b+4c<4a<2b+4c b+4c<4a<4b+4c b+3c<3a<3b+3c

Solution:

r + r1 = r2 + r3

r - r2 = r3 - r  

∆/s - ∆/s-b = ∆/s-c - ∆/s-a

- b/s(s-b) = c-a/(s-a)(s-c)

(s-a)(s-c)/ s(s-b) = a-c/b

tan B/2 = √(s-a)(s-c) / s(s-b)

tan² B/2 = a-c/b  --------------1

π/2 > B > π/3

π/4 > B/2 > π/6

1/3 < tan²(B/2) < 1

1/3 < a-c/b < 1

b/3 < a-c < b

b < 3a-3c

a-c < b

b + 3c < 3a