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The incircle of DeltaABC touches side BC...

The incircle of `DeltaABC` touches side BC at D. The difference between BD and CD (R is circumradius of `DeltaABC`) is (a) `|4 R sin.(A)/(2)sin.(B -C)/(2)|` (b) `|4R cos.(A)/(2) sin.(B -C)/(2)|` (c) `|b-c|` (d) `|(b-c)/(2)|`

A

`|4 R sin.(A)/(2)sin.(B -C)/(2)|`

B

`|4R cos.(A)/(2) sin.(B -C)/(2)|`

C

`|b-c|`

D

`|(b-c)/(2)|`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`|BD -CD| = |(s-b) - (s-c)|`
`= |b-c|`
`= 2R sin.(B-C)/(2)."cos"(B+C)/(2)`
`=|4R sin.(B-C)/(2)."sin"(A)/(2)|`
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