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Mathematics Form 3 Chapter 6 Angles and Tangents of Circles Mathematics Form 3 Chapter 6 Angles and Tangents of Circles
(b) ∠ACD = ∠ABD = 50° (c) x = 40° 5. x = ∠FEC = 50°
∠BCD = 50° + 60° = 110° y = 75° PT3 Standard Practice 6 ∠FOB = 360° – 78° – 90° – 50°
x = 180° – 110° = 142°
= 70° 9. (a) ∠CGF = 60° Section A 142°
(c) ∠ABC = 180° – 85° = 95° ∠FGE = 70° y = 2 = 71°
x = 180° – 40° – 95° x = 180° – 60° – 70° 1. ∠POQ = 180° – 35° – 35° = 110° ∴ x + y = 50° + 71° = 121°
= 45° = 50° ∠QOR = 180° – 63° – 63° = 54° Answer: D
x = 360° – 110° – 54°
5. (a) ∠DAE = ∠AEB = 30° (b) ∠SOT = 20° × 2 = 40° = 196° Section B

∠ADE = 180° – 30° – 70° ∠OTS = (180° – 40°) ÷ 2 = 70° Answer: B
∠QST = 110°
= 80° 1.
x = ∠ADE = 80° ∠QTS = 180° – 110° – 20° = 50° Statement
x = 70° – 50° = 20° 2. OQ = PQ – OP Pernyataan
= 8 – 5
(b) 2x + x = 180° (c) ∠EDF = 40° = 3 cm
3x = 180° ∠EDB = 60° + 40° = 100° (a) ∠OFE ≠ 90°
x = 60° QT = 5 – 3 2
2
∠ EFB = 180° – 100° = 80° (b) OB || PC ✓
y = 2 × 60° = 120° ∠DFB = 80° – 38° = 42° = 4 cm
∠DEB = ∠DFB = 42° = 40 mm (c) ∠k = ∠m ✓
(c) ∠PQR = 180° – 70° – 45° = 65° x = ∠DEB = 42° ∴ ST = 2 × QT = 80 mm
x = 180° – 65° = 115°
Answer: C (d) ∠m + ∠n = 180° ✓
6. (a) Yes
(b) No 3. ∠ODC = ∠OCD = y
(c) No = 180° – 100° 2. (a)
(d) Yes 2
(e) No y = 40° 30° x = 30°
x
∠ABC + ∠ADC = 180°
7. (a) x = 90° – 43° = 47° 120° + x + 40° = 180°
y = 43° x = 180° – 120° – 40°
x = 20° (b)
x + y = 90°
∴ x + y = 20° + 40° = 60° x O 30° x = 60°
(b) x = 90° Answer: C
y = 40°
4. ∠MNL = 68° – 35° = 33° (c)
(c) ∠ROQ = ∠POQ = 65°
x = 360° – 2(65°) = 230° x = 90° – 33° = 57° x O x = 100°
y = 90° – 65° = 25°
∠NOK = 68° × 2 100°
= 136°
8. (a) x = 60° 180° – 136°
y = 180° – 60° – 35° ∠NKO = 2 = 22° (d)
= 85° 50°
∠NKL = ∠MNL O x = 100°
22° + y = 33° x
(b) x = 70° y = 11°
y = 80° – 60°
= 20° ∴ y + x = 57° + 11° = 68°
Answer: A

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BOOKLET ANS MATH F3.indd 14 03/01/2020 10:20 AM