Solutions — Circle Theorems

1. Angle at the centre. Fluency

   • (a) Central 80°: APB = ½ × 80 = 40°. (Angle at circumference = half central angle, same arc.)
   • (b) Inscribed 55°: AOB = 2 × 55 = 110°. (Central angle = twice inscribed angle.)
   • (c) Central 150°: ACB (major arc) = ½ × 150 = 75°.
   • (d) Reflex central 240°: The non-reflex central angle = 360 − 240 = 120°. P is on the major arc (opposite side to the 240° arc). APB = ½ × 120 = 60°.

2. Angle in a semicircle. Fluency

   • (a) CAB=35°, angle in semicircle 90°: ACB = 90° (angle in semicircle). CBA = 180 − 90 − 35 = 55°.
   • (b) AB=10, angle ABC=40°: ACB = 90°. AC = AB × cos 40° = 10 × 0.766 ≈ 7.66 cm. (AC is adjacent to 40° in right triangle ACB.)
   • (c) DAB=52°: ADB = 90° (angle in semicircle). ABD = 180 − 90 − 52 = 38°.
   • (d) Triangle with sides 5, 12, 13: 5² + 12² = 25 + 144 = 169 = 13². So it is a right triangle with the right angle opposite the hypotenuse (diameter). Consistent with Thales’ theorem. ✓

3. Angles in the same segment. Fluency

   • (a) ADB=42°: ACB = 42°. (Angles in the same segment subtending arc AB are equal.)
   • (b) PRQ=67°: PSQ = 67°. (Same arc PQ, same segment.)
   • (c) (2x+10)=(4x−20): 2x+10 = 4x−20 ⇒ 30 = 2x ⇒ x = 15. Angle = 2(15)+10 = 40°.
   • (d) ACB=38°, D in minor segment: Central angle (minor arc) = 2×38 = 76°. Reflex central angle = 284°. D is in the minor segment, so ADB = ½×284° = 142°. (Alternatively: ADB = 180−38 = 142°, as opposite angles of a cyclic quadrilateral are supplementary.)

4. Cyclic quadrilaterals. Fluency

   • (a) A=82°: C = 180 − 82 = 98°.
   • (b) P=110°, Q=75°: R = 180 − 110 = 70°. S = 180 − 75 = 105°.
   • (c) (3x+5)+(2x−10)=180: 5x − 5 = 180 ⇒ 5x = 185 ⇒ x = 37. Angle A = 116°, angle C = 64°.
   • (d) Rectangle as cyclic quad: A rectangle has all angles 90°. Opposite angles: 90+90=180° ✓. A rectangle is always a cyclic quadrilateral, and its diagonals are diameters of the circumscribed circle.

5. Find angles from the diagram. Understanding

   • (a) Angle ACB: AB is a diameter, so angle ACB = 90°. (Angle in a semicircle.)
   • (b) Angle ADB: AB is a diameter, so angle ADB = 90°. (Angle in a semicircle.)
   • (c) Angle CAD: In triangle ACD: ACB = 90°, ACD = 35° (given), so angle CAD = 180 − 90 − 35 = 55°.
   • (d) Angle ABD: In triangle ABD: ADB = 90°, ABD = 25° (given as angle CBD, which equals angle ABD since both subtend arc AD... re-reading: angle CBD = 25° in triangle ABD with right angle at D). ABD = 25°. (Or: angle DAB = 180−90−25 = 65°.)

6. Alternate segment theorem. Understanding

   • (a) Tangent-chord angle 48°: TCB = 48°. (Alternate segment theorem: inscribed angle = tangent-chord angle in the alternate segment.)
   • (b) TCB=63°: Tangent-chord angle = 63°.
   • (c) Tangent at P, angle 55° with PQ: PRQ = 55°. (Alternate segment theorem.)
   • (d) Chord is a diameter: If the chord is a diameter, the inscribed angle in the alternate segment is 90° (angle in a semicircle). The alternate segment theorem then says the tangent makes a 90° angle with the diameter, i.e., the tangent is perpendicular to the radius at that point. This is consistent with Theorem 6.

7. Combining theorems. Understanding

   • (a) Central 100°, C major, D minor: ACB = ½×100 = 50° (major arc). ADB = ½×(360−100) = ½×260 = 130° (minor arc, reflex central angle).
   • (b) ABCD cyclic, BAC=CAD=28°: Angle BAD = 56°. Opposite angle BCD = 180 − 56 = 124°.
   • (c) Isosceles ABC inscribed, OA bisects angle BAC: Since AB=AC, the triangle is isosceles. The perpendicular bisector of BC passes through A and the centre O. The angle bisector of BAC from A is also the perpendicular bisector of BC (by symmetry of isosceles triangles). Hence OA ⊥ BC. □
   • (d) AB diameter, BAC=40°, DBA=30°: ACB = 90° (diameter). In triangle ACB: ABC = 180−90−40 = 50°. ABD = 30° (given). DBC = 50−30 = 20°. Angle ACD = angle ABD = 30° (same arc AD, same segment). (Or: angle ADB = 90°, so DAB = 60° ⇒ ACD = 60° by same arc... depends on position of D; answer 30° if D shares arc with A in the same way.)

8. Cyclic quadrilateral with algebra. Understanding

   • (a) 5x + (x+12) = 180: 6x + 12 = 180 ⇒ 6x = 168 ⇒ x = 28. Angle A = 140°, angle C = 40°.
   • (b) (3y+20)+(y+40) = 180: 4y + 60 = 180 ⇒ 4y = 120 ⇒ y = 30.
   • (c) Angles 2:3:4:5, total 360°: Parts: 2+3+4+5=14. Each part = 360/14. But opposite pairs must sum to 180. Check: 2k+5k=7k=180 ⇒ k=180/7; 3k+4k=7k=180 ✓. Angles: 2×180/7≈51.4°, 3×180/7≈77.1°, 4×180/7≈102.9°, 5×180/7≈128.6°.
   • (d) Square as cyclic quad: Yes, always. A square has all angles 90°. Opposite angles: 90+90=180° ✓. The circumscribed circle has its centre at the intersection of the diagonals and radius = half the diagonal length.

9. Angle chains. Problem Solving

   • (a) Angle BDA (BD is diameter): BD is a diameter, so angle BAD = 90° (angle in semicircle). In triangle ABD: BDA = 180 − 90 − 28 = 62°. (angle ABD = 28° given.)
   • (b) Angle BAD: 90° (angle in a semicircle, BD is diameter).
   • (c) Angle BCD (ABCD cyclic): Angle BAD = 90°. Opposite angle BCD = 180 − 90 = 90°. (Opposite angles of a cyclic quadrilateral sum to 180°.)
   • (d) Central angle BOD: BD is a diameter, so angle BOD (at centre) = 180°. This confirms BD is a diameter — the central angle subtended by a diameter is always 180°. ✓

10. Circle theorem proof. Problem Solving

    • (a) Triangle OAP (OA=OP=r, isosceles): Let angle OAP = angle OPA = α (base angles of isosceles triangle). Then angle AOP (exterior at O for triangle OAP with OP extended) = 2α. More precisely: angle AOQ = 2α where Q is the point on OP extended to the circle.
    • (b) Triangle OBP similarly: Let angle OBP = angle OPB = β. Then angle BOQ = 2β.
    • (c) Central angle AOB: Angle AOB = angle AOQ + angle BOQ = 2α + 2β = 2(α + β).
    • (d) Simplify: Angle APB = α + β (from angles in triangle APB at vertex P). Therefore angle AOB = 2(α+β) = 2 × angle APB. □