Published by:
CGP EDU Academic Team
Published on: August 13, 2026
The radius of the circle passing through the centre of the in-circle of Δ ABC and through the end points of BC is given by
A
cos A
B
sec A/2
C
sin A
D
a sec A/2
Text Solution
Verified by ExpertsThe correct answer is:
B
∠ BIC = 180 0 – 
So ∠ BOC = 2 
= 360 0 – (B + C)
= 180 0 – A { ∴ A + B + C = 180 0 )
= π – A
In Δ OBC
cos ( π – A) = 
⇒ a 2 = 2R 2 (1 + cos A)
R = 
= 
sec 
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