Question

ABCD is a rectangle, if ∠BPC = 124°
Calculate:

1. ∠BAP
2. ∠ADP

Answer 1

Diagonals of rectangle are equal and bisect each other.

∠PBC = ∠PCB = x (say)

But ∠BPC + ∠PBC + ∠PCB = 180°

124° + x + x = 180°

2x = 180° – 124°

2x = 56°

⇒ x = 28°

∠PBC = 28°

But ∠PBC = ∠ADP [Alternate ∠s]

∠ADP = 28°

Again ∠APB = 180° – 124°

∠APB = 56°

Also, PA = PB

∠BAP = 12 (180° – ∠APB)

= 12 × (180°- 56°)

= 12 × 124°

= 62°

Hence, (i) ∠BAP = 62° (ii) ∠ADP = 28°