The figure is not drawn to scale. PQT is an equilateral triangle and SQR is an isosceles triangle. RQP and QSU are straight lines and RP//UT. ∠TQU = 90° and ∠STQ = 52°.
- Find ∠RQU.
- Find ∠QSR.
- Find ∠TSU.
(a)
∠TQP = 60° (Equilateral triangle PQY)
∠TQU = 90°
∠STQ = 52°
∠RQU
= 180° - ∠TQU - ∠TQP
= 180° - 90° - 60°
= 30° (Angles on a straight line)
(b)
∠QSR
= (180° - ∠RQU ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(c)
∠TSQ
= 180° - ∠TQU - ∠STQ
= 180° - 90° - 52°
= 38°
∠TSU
= 180° - ∠TSQ
= 180° - 38°
= 142° (Angles on a straight line)
Answer(s): (a) 30°; (b) 75°; (c) 142°
