In the figure, O Is the centre of the circle and SX is parallel to TU. VY = VX, ∠OTS = 54° and ∠YXV = 53°. Find
- ∠UTZ
- ∠TUV
(a)
OS = OT = Radius
∠STO = ∠TSO = 54° (Isosceles triangle, OSB)
∠TOZ
= 54° + 54°
= 108° (Exterior angle of a triangle)
OT = OZ = Radius
∠OZT
= (180° - 108°) ÷ 2
= 72° ÷ 2
= 36°
∠UTZ = 36° (Alternate angles, TU//SE)
(b)
VY = VX
∠VXY = ∠VYX = 53° (Isosceles triangle VXF)
∠XVY
= 180° - 53° - 53°
= 74°
∠ZVU = ∠XVY = 74° (Vertically opposite angles)
∠TUV
= 180° - 74°
= 106° (Interior angles)
Answer(s): (a) 36°; (b) 106°