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 = 50° and ∠WTX = 34°. Find
- ∠QTS
- ∠TVU
(a)
∠QTX
= (180° - 50°) ÷ 2
= 65° (Isosceles triangle)
∠QTS
= 180° - 65°
= 115° (Angles on a straight line)
(b)
∠VTU
= 65° - 34°
= 31°
∠TVU
= (180° - 31°) ÷ 2
= 74.5 ° (Isosceles triangle)
Answer(s): (a) 115°; (b) 74.5°