Question

in triangle ABC,P,Q are two points on side AC such that AB=AQ,BC=PC angleABC=76 degree find angle PBQ​

Answer 1

Given :  In triangle ABC,P,Q are two points on side AC such that AB=AQ,BC=PC angleABC=76 degree

To find : angle PBQ

Solution:

in Δ ABQ

AB = AQ

=> ∠ABQ = ∠AQB

∠ABQ + ∠AQB + ∠A = 180°

=> 2∠ABQ   + ∠A = 180°

Similarly in Δ CBP

BC = PC

2∠CBP   + ∠C = 180°  

Adding both

2∠ABQ  + 2∠CBP + ∠A  +  ∠C = 180°  + 180°

 ∠A  +  ∠C = 180° - ∠B     ∠B  = ∠ABC = 76°

=> ∠A  +  ∠C = 180° - 76°

=> 2∠ABQ  + 2∠CBP +180° - 76° = 180°  + 180°  

=>  2∠ABQ  + 2∠CBP = 256°

=>  ∠ABQ  +  ∠CBP  = 128°

∠ABQ  +  ∠CBP = ∠B + ∠PBQ

=> 128° = 76° + ∠PBQ

=> ∠PBQ  = 52°

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