In the figure, QRST is a parallelogram. O is the centre of the circle. Find
- ∠QTS
- ∠QRT
(a)
∠ORT = ∠RTO = 27° (Angle properties within a circle)
∠TOR
= 180° - 27° - 27°
= 126° (Angles sum of triangle)
∠OQR + ∠ORQ = ∠ROT (Exterior angle of a triangle)
∠OQR = ∠ORQ (Isosceles triangle, OQ = OR)
∠OQR
= 126° ÷ 2
= 63° (Exterior angle of a triangle)
∠QTS
= 180° - 63°
= 117° (Interior angles)
(b)
∠QRT
= 180° - 63° - 27°
= 90° (Angle sum of triangles)
Answer(s): (a) 117°; (b) 90°