In the figure, QR = QT, ∠RQT = 35° and ∠UST = 15°. Find
- ∠q
- ∠r
(a)
∠QRT
= (180° - 35°) ÷ 2
= 72.5° (Isosceles triangle)
∠q
= 180° - 15° - 72.5°
= 92.5° (Angles sum of triangle)
(b)
∠QTS
= 180° - 72.5°
= 107.5° (Angles on a straight line)
∠r
= 15° + 107.5°
= 122.5° (Exterior angle of a triangle)
Answer(s): (a) 92.5°; (b) 122.5°