In the adjoining figure, PS is the bisector of ∠QPR and PT ⊥ QR and ∠Q > ∠R, then ∠TPS is -

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Rajasthan 3rd Grade (Maths & Science) Official Paper (Held On: 25 Feb 2023 Shift 2)
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  1. \(\rm 90^\circ-\frac{1}{2}(\angle Q-\angle R)\)
  2. \(\rm \frac{1}{2}(\angle Q+\angle R)\)
  3. ∠Q - ∠R
  4. \(\rm \frac{1}{2}(\angle Q-\angle R)\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{1}{2}(\angle Q-\angle R)\)
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Detailed Solution

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Given

PS is the bisector of angle QPR

PT is perpendicular to QR

Solution

In the ΔPQR,

∠RPS = ∠QPS = yº and ∠TPS = xº 

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In ΔPTR,

⇒ ∠(x + y)º + 90º + ∠R = 180  ..(1)

In ΔPTQ,

⇒ ∠(y - x)º + 90º + ∠Q = 180 ..(2)

Equating both equations,

⇒ ∠(x + y)º + 90º + ∠R = ∠(y - x)º + 90º + ∠Q

⇒ 2∠xº = ∠Q - ∠R

⇒ ∠xº = (∠Q - ∠R)/2

The correct option is 4.

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