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

hintHint:

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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