Question

#15 POINTS + 1 FOLLOWER.

IF YOU ANSWER MY QUESTION..

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Answer 1

Solution :-

Given : ABC is a triangle in which ∠A = 72°.

The internal bisectors of angles B and C meet in O.

∠BAC + ∠ABC + ∠ACB = 180° [Angle side property of a △]

=> 72° + ∠ABC + ∠ACB = 180°

=> ∠ABC + ∠ACB = 180° - 72°

=> ∠ABC + ∠ACB = 108°

=> ∠ABC/2 + ∠ACB/2 = 108°/2 [Dividing by 2]

=> ∠ABC/2 + ∠ACB/2 = 54°

Now,

∠ABC/2 + ∠ACB/2 + ∠BOC = 180° [Angle side property of a △]

=> 54° + ∠BOC = 180° [Substituting the value]

=> ∠BOC = 180° - 54° = 126°

Hence, ∠BOC = 126°

Answer 2

SOLUTION

In ∆ABC

∠A + ∠B +∠C=180°(By angle sum prop.)

72°+ ∠B+ ∠C= 180°

=) ∠B+ ∠C= (180-72)°

=) ∠B + ∠C= 108°.........(1)

In ∆BOC

∠B+∠C+∠O= 180°(by angle sum prope.)

∠BOC+ 1/2∠C+ 1/2∠B= 180°

(since BO & CO are bisectors of ∠B & C respectively.

∠BOC + 1/2(∠B + ∠C)=180°

=)∠BOC+ 1/2× 108°= 180°

=) ∠BOC= 180°- 54°

=) ∠BOC= 126°

hope it helps ☺️