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Question
ABCD is a rectangle, if ∠BPC = 124°
Calculate:
- ∠BAP
- ∠ADP
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Solution
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°
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