In the figure, QRST is a trapezium and QTUX is a rhombus. QT is parallel to RS and TU = TV. XTS and TWV are straight lines. ∠TQX = 51° and ∠WTX = 28°. Find
- ∠QTS
- ∠TVU
(a)
∠QTX
= (180° - 51°) ÷ 2
= 64.5° (Isosceles triangle)
∠QTS
= 180° - 64.5°
= 115.5° (Angles on a straight line)
(b)
∠VTU
= 64.5° - 28°
= 36.5°
∠TVU
= (180° - 36.5°) ÷ 2
= 71.75 ° (Isosceles triangle)
Answer(s): (a) 115.5°; (b) 71.75°