The bisectors of exterior angles at `B\ a n d\ C` of `/_A B C` meet at `Odot` If `/_A=x^0,` then `/_B O C=` `90^0+(x^0)/2` (b) `90^0-(x^0)/2` `180^0+(x^0)/2` (d) `180^0-(x^0)/2`

Explanation

`/_OBC = 180@ - /_B - 1/2 (180°@- /_B)
/_OBC = 90@ - 1/ 2 (/_B )
And, /@OCB = 180@ - /_C - 1 /2 (180@- ∠C)
∠OCB = 90@ - 1/2∠C
In ΔBOC, ∠BOC + ∠OCB + ∠OBC = 180°
∠BOC + 90° -1/2∠C + 90° - 1/2∠B = 180°
∠BOC = 1/2(∠B + ∠C)
∠BOC = 1/2(180° - ∠A)
[From ] ∠BOC = 90° - 1/2(∠A) ∠BOC = 90@ - x/ 2@`