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 ?

15

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°