Question

In a triangle ABC, the internal bisectors of angle B and angle C meet at P and the external bisectors of angle B and angle C meet at Q, then angle P = angle Q = 60 angle P = angle Q = 90 angle P = angle Q = 45 none of these

Answer 1

Answer:

Angle P=Angle Q=90 degrees

Step-by-step explanation:

∠ACB and ∠QCB form a linear pair,

So, ∠ACB + ∠QCB = 180º

=> ∠ACB/2 + ∠QCB/2 = 180º /2

=> ∠ACB/2 + ∠QCB/2 = 90º

Again since PC and QC are the angle bisectors

=> ∠PCB + ∠BCQ = 90º

=> ∠PCQ = 90º

Similarlry,

∠PBQ = 90º

In qradrilateral BPCQ,

Sum of all four angles = 360º

=> ∠BPC + ∠PCQ + ∠CQB + ∠QBP = 360º

=> ∠BPC + ∠BQC + 180º = 360º

=> ∠BPC + ∠BQC = 360º - 180º

=> ∠BPC + ∠BQC = 180º

=> ∠P = ∠Q = 90º [Since ∠PCQ = ∠PBQ = 90º]

Hope it helps.

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