The incircle of `DeltaABC` touches side BC at D. The difference between BD and CD (R is circumradius of `DeltaABC`) is

Question

The incircle of `DeltaABC` touches side BC at D. The difference between BD and CD (R is circumradius of `DeltaABC`) is

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)|`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 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)|`

5 views Answered 2023-02-05 15:29