In the figure, O Is the centre of the circle and SX is parallel to TU. VY = VX, ∠OTS = 56° and ∠YXV = 55°. Find
- ∠UTZ
- ∠TUV
(a)
OS = OT = Radius
∠STO = ∠TSO = 56° (Isosceles triangle, OSB)
∠TOZ
= 56° + 56°
= 112° (Exterior angle of a triangle)
OT = OZ = Radius
∠OZT
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠UTZ = 34° (Alternate angles, TU//SE)
(b)
VY = VX
∠VXY = ∠VYX = 55° (Isosceles triangle VXF)
∠XVY
= 180° - 55° - 55°
= 70°
∠ZVU = ∠XVY = 70° (Vertically opposite angles)
∠TUV
= 180° - 70°
= 110° (Interior angles)
Answer(s): (a) 34°; (b) 110°
