Question

in the given rectangle ABCD (fig 2),BP=BC.what is the measure angle PCD(isska answer hai 45 hai plz pura links)

Answer 1

ABCD is a rectangle
So all the angles are 90°
So,
Angle PBC = 90° and BCD = 90°
And Δ PBC is an isocels triangle
So angle BPC = BCP
Angle ( BPC + BCP + PBC ) = 180°
Angle ( BPC + BPC + PBC ) = 180°
Angle ( 2BPC + 90° ) = 180°
Angle ( 2BPC ) =180° -90°
Angle ( BPC ) = 90°/2
Angle BPC = 45°
So angle BCP = 45°
Angle PCD = 90° - Angle BCP
Angle PCD = 90°- 45°
Angle PCD = 45°

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Answer 2

abcd is a rectangle

let we take all the angles be 90°

angle PBC= 90° and BCD = 90°

ANGLE BPC + BCP + PBC

ANGLE BPC + BPC + PBC= 180°

ANGLE 2BPC + 90°=180°

angle 2BPC =180°-90°

angle 2bpc = 90°

angle BPC = 90°

angle BPC =90°/2

angleBPC =45°