ΔPQR is inscribed in a circle with centre O. PO is produced to meet QR at U and the circle at S, and PT ⊥ QR, where T lies between Q and U. If ∠Q = 70° and ∠R = 55°, then what is the measure (in degrees) of ∠TPS ?
Answer & explanation
Correct answer: option 3
In right angles triangle ΔPTQ
∠PQT + ∠QPT + ∠QTP = 180°
∠QPT = 180° - (70° + 90°)
∠QPT = 20°
In ΔPQR
∠P + ∠Q + ∠R = 180°
∠P = 180° – (70° + 55°)
∠P = 55°
Property used here is :- [Angle at centre are doubled of angle at circumference]
∠POR = 2 × ∠PQR
∠POR = 2 × 70°
∠POR = 140°
Now,
In ΔPOR
PO = OR ( Because radius of circle)
Property used here is :- [Angle opposite to equal sides are equal in length]
∠OPR = ∠ORP
∠OPR = \(\frac{180° – ∠POR}{2}\)
∠OPR = \(\frac{180° – 140°}{2}\)
∠OPR = 20°
Then,
∠TPS + (∠QPT + ∠OPR) = ∠QPR
∠TPS = 55° – (20° + 20°)
∠TPS = 55° – 40°
∠TPS = 15°