Question

In triangle ABC angle A =50 and the external bisectors of angle b and angle c meet at o . The measure of angle boc is

Answer 1

In ΔABC , The measure of angle BOC is 65°  where ∠A = 50° and external bisectors of ∠B and ∠C meet at O

Given :  ΔABC , ∠A = 50° ,

the external bisectors of ∠B and ∠C meet at O  

To find : measure of  ∠BOC  

Solution:

Exterior angle = Sum of opposite interior angles

Exterior angle of B is ∠CBX

Exterior angle of C is ∠BCY

∠CBO is bisector of ∠CBX

∠BCO is bisector of ∠BCY

∠CBX  = ∠C + ∠A

∠BCY = ∠B + ∠A

∠CBO = (1/2)∠CBX  = (1/2) (∠C + ∠A)

∠BCO = (1/2)∠BCY  = (1/2) (∠B + ∠A)

∠CBO  + ∠BCO  + ∠BOC = 180°   ( Sum of angles of a triangle)

=>  (1/2) (∠C + ∠A) +  (1/2) (∠B + ∠A)  +  ∠BOC = 180°

=> (1/2) (∠C + ∠A + ∠B) +  (1/2) (∠A)  +  ∠BOC = 180°

=> (1/2) (180°) +  (1/2) (∠A)  +  ∠BOC = 180°

=> 90°  +  (1/2) (∠A)  +  ∠BOC = 180°

=> ∠BOC = 90° -  (1/2) (∠A)  

∠A = 50°

=> ∠BOC = 90° -  (1/2) ( 50°)  

=> ∠BOC = 90° -  25°

=> ∠BOC =65°  

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BOC = 104° , ∠OAC = 14° ∠BAC = 52

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