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Maths-

General

Easy

Question

In $straight triangle A B C$ , the bisectors of $straight angle A B C$ and $straight angle B C A$ intersect each other at O. Prove that $straight angle B O C equals$ $90 to the power of ring operator plus 1 half straight angle A$

hint Hint:

As we know OB and OC are angular bisectors we find ∠OBC and ∠OCB in terms of ∠A and ∠B .We find the value of ∠BOC by using the sum of angles in the triangle is 180°.

The correct answer is: Hence proved

Aim :- Prove $straight angle B O C equals 90 plus 1 half straight angle A$

Step 1:- Find ∠OBC and ∠OCB
∠OBC = $1 half$ ∠B(OB is the angular bisector)
∠OCB = $1 half$ ∠C (OC is the angular bisector)
∠A+ ∠B+ ∠C =180° (sum of angles in triangle ABC)
∠B+ ∠C =180°-∠A
Step 2:- Find ∠BOC
∠BOC+ ∠OBC + ∠OCB =180° (sum of angles in triangle OBC)
∠B+ ∠C =180°-∠A
Step 2:- Find ∠BOC
∠BOC+ ∠OBC + ∠OCB =180° (sum of angles in triangle OBC)
∠BOC+ $1 half$ ∠B + $1 half$ ∠C = 180°
∠BOC= 180° -( $1 half$ ∠B +∠C )
∠BOC= 180°-( $1 half$ ∠B +∠C)
∠BOC= 180°-( $1 half$ 180°-∠A)
$not stretchy rightwards double arrow straight angle B O C equals 90 to the power of ring operator plus 1 half straight angle A$
Hence proved

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