Question

In ∆ ABC, BO and CO are bisectors of ∠ABC and ∠ACB. Prove that ∠BOC=90 degree+ 1⁄2∠A.

Answer 1

In ΔABC, by angle sum property we have
2x + 2y + ∠A = 180°
⇒ x + y + (∠A/2) = 90°
⇒ x + y = 90° –  (∠A/2)  à (1)
In ΔBOC, we have
 x + y + ∠BOC = 180°
90° – (∠A/2) + ∠BOC = 180° [From (1)]
∠BOC = 180° – 90° + (∠A/2)
∠BOC = 90° + (∠A/2)

Answer 2

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Mate :-

Solution :-

☆ In ∆ ABC , We have

∠A + ∠ B + ∠C = 180

:- 1/2∠A + 1/2∠ B +1/2 ∠C = 1/2×180

=> 1/2∠A + ∠ 1 + ∠2 = 90

=> ∠ 1 + ∠2 = 90 - 1/2∠A.....1 eq

Now, in ∆BOC We have

∠ 1 + ∠2 + ∠BOC = 180

( 90 - 1/2∠A ) + ∠BOC = 180 (using. ..1 )

=> BOC=90 degree+ 1⁄2∠A.