Question

7th question (1stpic)
2nd question separately

Answer 1

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$<b><u>Answer</u></b>$

(1) Since, PBCR becomes rectangle.

∴ PB = RC = x

=> AP = x

If QC = RC = x

Since, ∠PRC = 90°

∴ ∠PRQ = 45°

And some of the angles of a triangle is 180°.

∴ ∠PQR = 90° and ∠PRQ = 45°

∴ ∠QPR = 180° - (90° + 45°)

∴ ∠QPR = 45°

Hence, the proof.

(2) In △PBC, we have,

∠P + ∠PBC + ∠C = 180° (angle sum property of a triangle)

$∠P + \frac{1}{2} B + C = 180°$ => eq. (1)

In△QAD, we have,

∠Q + ∠A + ∠ADQ = 180° (angle sum property)

$∠Q + ∠A + \frac{1}{2} D = 180°.$ => eq. 2.

But, ∠A + ∠B + ∠C + ∠D = 360°

$∠P + ∠Q + ∠A + ∠C \frac{1}{2} (∠B + ∠D) = ∠A + ∠B + ∠C + ∠D$

∠P + ∠Q = $\frac{1}{2} (∠B + ∠D)$

∠P + ∠Q = $\frac{1}{2} (∠ABC + ∠ADC)$

Thanks....!

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