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In triangleABC, tanA/2=5/6, tanC/2=2/5, ...

In `triangleABC, tanA/2=5/6, tanC/2=2/5`, then

A

`b^(2)=ac`

B

`2b=ac`

C

`2ac=b(a+c)`

D

a+b+c

Text Solution

Verified by Experts

We have,
`tanA/2=5/6` and `tanC/2=2/5`
`rArrtanA/2tanC/2=1/3`
`rArr(Delta)/(s(s-a))xx(Delta)/(s(s-c))=1/3`
`rArr(s-b)/s=1/3rArr2s=3brArra+c=2b`
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