In the figure, O Is the centre of the circle and SX is parallel to TU. VY = VX, ∠OTS = 53° and ∠YXV = 56°. Find
- ∠UTZ
- ∠TUV
(a)
OS = OT = Radius
∠STO = ∠TSO = 53° (Isosceles triangle, OSB)
∠TOZ
= 53° + 53°
= 106° (Exterior angle of a triangle)
OT = OZ = Radius
∠OZT
= (180° - 106°) ÷ 2
= 74° ÷ 2
= 37°
∠UTZ = 37° (Alternate angles, TU//SE)
(b)
VY = VX
∠VXY = ∠VYX = 56° (Isosceles triangle VXF)
∠XVY
= 180° - 56° - 56°
= 68°
∠ZVU = ∠XVY = 68° (Vertically opposite angles)
∠TUV
= 180° - 68°
= 112° (Interior angles)
Answer(s): (a) 37°; (b) 112°