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 = 47° and ∠WTX = 33°. Find
- ∠QTS
- ∠TVU
(a)
∠QTX
= (180° - 47°) ÷ 2
= 66.5° (Isosceles triangle)
∠QTS
= 180° - 66.5°
= 113.5° (Angles on a straight line)
(b)
∠VTU
= 66.5° - 33°
= 33.5°
∠TVU
= (180° - 33.5°) ÷ 2
= 73.25 ° (Isosceles triangle)
Answer(s): (a) 113.5°; (b) 73.25°