In ΔPQR, QT ⊥ PR and S is a point on QR such that ∠PSQ = p°. If ∠TQR = 46° and ∠SPR = 32°, then the value of p is∶
76°
Suppose QT and PS intersecting at point O.
Given, ∠PSQ = P°, and ∠TQR = 46°
∠SPR = 32° and ∠QTP = 90°
In ΔPOT,
⇒ ∠P + ∠O + ∠T = 180
⇒ ∠O = 180° – 32° – 90° = 58°
⇒ ∠POT = ∠QOS = 58° (Vertically opposite angle)
In ΔQOS
⇒ ∠O + ∠Q + ∠S = 180°
⇒ P° = 180° – 46° – 58° = 76°
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