Question

Bisector of angle A and AngleB of a triangle abc intersect each other at o then find Angle BOC​

Answer 1

Answer:

Given:-A traingle ABC in which BO and CO are the bisectors of <B and <C respectively

To prove:-<BOC-(90 +1/2<A)

Proof:

We know that the sum of three angled of a triangle is 180°

<A+<B+<C=180°

1/2<A+1/2<B+1/2<C-90°

1/2<A+<OBC+<OCB-90°

<OBC+<OCB=(90°-1/2<A)

Now,in traingle ABC

<OBC+<OCB+<BOC-180°

(90-1/2<A)+<BOC=180°

<BOC=180°-(90°-1/2<A)

(90°+1/2<A)

Hence,<BOC=(90*+1/2<A)

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Answer 2

Answer:

∠ABO=∠OBC=∠ABC/2

∠ACO=∠OCB=∠ACB/2

In△ABC,

∠ABC+∠ACB+∠BAC=180∘

⟹∠ABC+∠ACB=180∘−∠BAC

Dividing both sides by2,

⟹∠ABC+∠ACB/2=90∘−∠BAC/2...(1)

In△BOC,

∠ABC/2+∠ACB/2+∠BOC=180∘

⟹∠ABC+∠ACB/2+∠BOC=180∘

⟹∠BOC=180∘−∠ABC+∠ACB/2

From(1),

⟹∠BOC=180∘−[90∘−∠BAC/2]

⟹∠BOC=180∘−90∘+∠BAC/2

⟹∠BOC=90∘+∠BAC/2