PQRS is a square. GFR and QFS are straight lines. QP = QG and ∠FQG = 15°. Find
- ∠QPG
- ∠SRG
.
(a)
∠PQS = 45° (Right angle)
∠PQG
= ∠PQS - ∠FQG
= 45° - 15°
= 30°
∠QPG
= (180° - ∠PQG) ÷ 2
= (180° - 30°) ÷ 2
= 150° ÷ 2
= 75° (Isosceles triangle)
(b)
QP = QG = QR
GQR is an isosceles triangle.
∠QGR = ∠QRG (Isosceles triangle)
∠QRG
= (180° - ∠SQR - ∠FQG) ÷ 2
= (180° - 45° -15°) ÷ 2
= 120° ÷ 2
= 60°
∠SRG
= ∠QRS - ∠QRG
= 90° - 60°
= 30°
Answer(s): (a) 75°; (b) 30°
