In the figure, O Is the centre of the circle and SX is parallel to TU. VY = VX, ∠OTS = 52° and ∠YXV = 53°. Find
- ∠UTZ
- ∠TUV
(a)
OS = OT = Radius
∠STO = ∠TSO = 52° (Isosceles triangle, OSB)
∠TOZ
= 52° + 52°
= 104° (Exterior angle of a triangle)
OT = OZ = Radius
∠OZT
= (180° - 104°) ÷ 2
= 76° ÷ 2
= 38°
∠UTZ = 38° (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) 38°; (b) 106°