Question

♡CHALLENGE :

❥The internal bisectors of angles B and C of a triangle ABC meet at O. Then, ∠BOC is equal to :

$(a) \: 90° + A$


$(b) \: 2A$


$(c) \: 90° + \frac{1}{2}A$


$(d) \: 180° - A$​

Answer 1

Answer:

its a theorem BOC = 90+ A/2

if it bisects ecternally then BOC= 90-A/2

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