Question 145):
ABCD is a rectangle, if ∠BPC = 124°
Calculate : (i) ∠BAP (ii) ∠ADP

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$∠ B A P = 26 °; ∠ A D P = 82 °$

$∠ B A P = 28 °; ∠ A D P = 62 °$

$∠ B A P = 52 °; ∠ A D P = 38 °$

$∠ B A P = 62 °; ∠ A D P = 28 °$

answer is B.

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Detailed Solution

1055226

Diagonals of rectangle are equal and bisect each other.

Let, ∠PBC = ∠PCB = x

But ∠BPC + ∠PBC + ∠PCB = 180°

124° + x + x = 180°

2x = 180° – 124°

2x = 56°

=> x = 28°

∠PBC = 28°

But ∠PBC = ∠ADP (Alternate angle)

So, ∠ADP = 28°

Again ∠APB = 180° – 124° = 56°

Also PA = PB

$∠ B A P = \frac{1}{2} (180^{\circ} - ∠ A P B)$

$= \frac{1}{2} (180^{\circ} - 56^{\circ}) = \frac{1}{2} \times 124^{\circ} = 62^{\circ}$

Hence (i) ∠BAP = 62° (ii) ∠ADP =28°.

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