Circle Theorems — Formal Proofs – Grade 10 IGCSE Maths

Welcome to Circle Theorems!

Circle theorems are powerful geometric results that let you find missing angles in circles without measuring. At Grade 10 IGCSE, you must not only use these theorems but also prove them and state them explicitly when answering questions. Missing the theorem name costs marks.

7 Key Theorems · Formal Proofs · Algebraic Problems · Exam Technique

Golden rule: In every exam answer, state which theorem you are using at each step. Examiners expect: "Angle at centre = 2 × angle at circumference" not just a number.

The 7 Theorems

All theorems with formal proofs

Multi-Step Problems

Chaining theorems together

Algebraic Problems

Set up equations using theorems

Worked Examples

6 fully solved problems

Theorem Explorer

Interactive circle diagrams

Formula Sheet

All 7 theorems at a glance

Learn 1 — The 7 Theorems with Formal Proofs

Theorem 1 — Angle at Centre = 2 × Angle at Circumference

Let O be the centre, A and B on the circle, C a point on the major arc. We want to prove ∠AOB = 2 × ∠ACB.

Step 1: Draw OC (a radius). Now OA = OC = OB (all radii), so △OCA and △OCB are both isosceles.

Step 2: Let ∠OCA = α and ∠OCB = β. Since △OCA is isosceles (OA = OC), we have ∠OAC = α too. By angle sum in △OCA: ∠AOC = 180° − 2α.

Step 3: Similarly in △OCB: ∠BOC = 180° − 2β.

Step 4: Angles around O sum to 360°, so:
∠AOB = 360° − (180° − 2α) − (180° − 2β) = 360° − 180° + 2α − 180° + 2β = 2α + 2β = 2(α + β)

Step 5: But ∠ACB = α + β (angles OCA and OCB), so ∠AOB = 2 × ∠ACB ✓

Theorem 2 — Angle in a Semicircle = 90°

This is a special case of Theorem 1 where AOB is a diameter.

Proof: Since AOB is a straight line (diameter), ∠AOB = 180°. By Theorem 1: ∠ACB = ½ × 180° = 90°. ✓

If AB is a diameter and C is any point on the circle (not A or B), then ∠ACB = 90°. Conversely, if ∠ACB = 90° then AB must be a diameter.

Theorem 3 — Angles in the Same Segment are Equal

Points C and D are both on the same arc, both looking at chord AB.

Proof: By Theorem 1, ∠AOB = 2 × ∠ACB and also ∠AOB = 2 × ∠ADB (same arc AB, same centre angle). Therefore ∠ACB = ∠ADB. ✓

Any two inscribed angles that subtend the same arc (same chord, same side) are equal. "Angles in the same segment" — both angles are in the same region of the circle.

Theorem 4 — Opposite Angles of a Cyclic Quadrilateral Sum to 180°

ABCD is a cyclic quadrilateral (all four vertices on the circle). Prove ∠A + ∠C = 180°.

Step 1: Let ∠BOD (reflex) = 2∠BAD (Theorem 1, ∠BAD subtends arc BCD).

Step 2: Let ∠BOD (non-reflex) = 2∠BCD (Theorem 1, ∠BCD subtends arc BAD).

Step 3: Reflex ∠BOD + non-reflex ∠BOD = 360° (angles at a point).
∴ 2∠BAD + 2∠BCD = 360° ⟹ ∠BAD + ∠BCD = 180° ✓

Theorem 5 — Tangent is Perpendicular to Radius

Line T is tangent to the circle at point P. O is the centre. Prove OP ⊥ T.

Proof by contradiction: Assume OP is NOT perpendicular to T. Then there exists a point Q on T such that OQ ⊥ T and OQ < OP. But then Q lies inside the circle (OQ < radius OP). A tangent by definition touches the circle at exactly one point and lies outside — Q being inside the circle contradicts this. Therefore our assumption was wrong, and OP must be perpendicular to T. ✓

Theorem 6 — Tangents from an External Point are Equal

From external point P, draw tangents PA and PB (A and B are points of tangency). Prove PA = PB.

Step 1: OA ⊥ PA and OB ⊥ PB (Theorem 5), so ∠OAP = ∠OBP = 90°.

Step 2: In triangles OAP and OBP: OA = OB (radii), OP = OP (shared hypotenuse), ∠OAP = ∠OBP = 90°.

Step 3: By RHS congruence, △OAP ≅ △OBP. Therefore PA = PB ✓

Also from this congruence: ∠APO = ∠BPO (OP bisects angle P) and ∠AOP = ∠BOP (OP bisects angle at centre).

Theorem 7 — Alternate Segment Theorem

The angle between a tangent and a chord equals the inscribed angle in the alternate segment.

Setup: Let T be a tangent at A, and AB a chord. Let C be a point in the alternate segment. We prove ∠(tangent, chord AB) = ∠ACB.

Step 1: Let ∠CAB = θ (angle in the alternate segment).

Step 2: Draw diameter AD. Since ∠ABD = 90° (angle in semicircle, Theorem 2), and ∠DAT = 90° (tangent ⊥ radius), we can relate the angles using complementary angles to establish ∠(tangent, AB) = θ. ✓

In plain language: The angle between the tangent and the chord equals the angle "on the other side" — the angle subtended by the same chord in the opposite segment.

Learn 2 — Multi-Step Problems

The 4-Step Method for Multi-Step Circle Problems:
1. Label ALL unknown angles with letters (a, b, c…) at the start
2. Identify every geometric feature: radii (→ isosceles triangles), diameters, tangents (→ right angles), parallel lines
3. Apply theorems one at a time, stating the theorem at each step
4. Chain results: use each answer as input for the next theorem

Spotting Which Theorem to Use

Radii present? → Look for isosceles triangles → base angles equal
Diameter present? → Angle in semicircle = 90°
Tangent present? → Tangent ⊥ radius (right angle); tangents from external point equal; alternate segment theorem
All 4 points on circle? → Cyclic quadrilateral → opposite angles sum to 180°
Two angles, same arc? → Angles in same segment are equal
Angle at centre vs circumference? → Centre = 2 × circumference

Example: Chain of Three Theorems

Problem: O is the centre. PA and PB are tangents from P. ∠APB = 48°. Find ∠AOB and then find the angle in the alternate segment at any point C on the major arc.

Step 1: OA ⊥ PA and OB ⊥ PB (Theorem 5) → ∠OAP = ∠OBP = 90°
Step 2: In quadrilateral OAPB: angles sum to 360°
∠AOB + 90° + 90° + 48° = 360° → ∠AOB = 132°
Step 3: ∠ACB = ½ × ∠AOB = ½ × 132° = 66° (Theorem 1: angle at centre = 2 × angle at circumference)

Key Alert: Reflex Angles at Centre

When the angle at the centre is reflex (greater than 180°), the formula still holds but you must use the reflex angle. If ∠AOB (reflex) = 240°, then the angle at the circumference on the minor arc side = 120°. Draw a diagram and mark which arc you're working with.

Isosceles Triangles — Don't Forget!

Any time two sides of a triangle are radii, the triangle is isosceles. This gives you equal base angles. This is one of the most commonly missed steps in multi-theorem problems.

Example: OA = OB = OC (radii). ∠OAB = 35°. Since △OAB is isosceles, ∠OBA = 35° too. Then ∠AOB = 180° − 35° − 35° = 110°.

Learn 3 — Algebraic Problems and Exam Technique

Setting Up Equations

Method: Express angles in terms of x using given information, then apply the appropriate theorem to form an equation.

Example 1: Angle at centre = 6x. Angle at circumference = 2x + 15.
Theorem 1: 6x = 2(2x + 15) → 6x = 4x + 30 → 2x = 30 → x = 15°
Check: centre = 90°, circumference = 45° ✓ (90 = 2 × 45)

Example 2: Cyclic quadrilateral. Two opposite angles are (3x + 10)° and (5x − 30)°.
Theorem 4: (3x + 10) + (5x − 30) = 180 → 8x − 20 = 180 → 8x = 200 → x = 25°

Proving a Point Lies on a Circle (Converse Theorems)

Converse of Theorem 2: If ∠ACB = 90° where C is a point and AB is a line segment, then C lies on the circle with diameter AB.

Converse of Theorem 4: If opposite angles of a quadrilateral sum to 180°, the quadrilateral is cyclic (can be inscribed in a circle).

These converses allow you to PROVE that certain configurations must be circular.

Common Exam Setups

Setup A — Tangent from external point: Find PA given OP = 13 cm, radius = 5 cm.
∠OAP = 90° (Theorem 5) → PA² = OP² − OA² = 169 − 25 = 144 → PA = 12 cm

Setup B — Two chords intersecting: Chords AC and BD intersect at E inside the circle.
∠AEB = ∠DEC (vertically opposite). Also ∠A = ∠D (same segment for arc BC). So △AEB ~ △DEC.

Setup C — Mixed diagram: Tangent-radius right angle combined with isosceles triangle from centre, then alternate segment applied. Label everything, chain three theorems.

Exam Technique: Writing Up Proofs

Every step needs a reason. Format:
∠ABC = 90° [angle in semicircle, diameter AC]
∠OAT = 90° [tangent perpendicular to radius at A]
∠XYZ = 2 × ∠XAZ [angle at centre = 2 × angle at circumference, arc XZ]

Reasons in square brackets score the method marks. Losing reasons = losing 1–2 marks per step.

Example 1 — Basic Theorem: Angle at Centre

O is the centre of a circle. A, B, C are points on the circle. ∠AOB = 112°. Find ∠ACB.

Identify: ∠AOB is the angle at the centre; ∠ACB is the angle at the circumference, both subtending arc AB.

Apply Theorem 1: Angle at centre = 2 × angle at circumference
112° = 2 × ∠ACB → ∠ACB = 56°

Answer: ∠ACB = 56° [angle at centre = 2 × angle at circumference]

Example 2 — Algebra with Theorems (solve for x)

Cyclic quadrilateral ABCD. ∠DAB = (4x + 5)° and ∠BCD = (2x + 35)°. Find x and both angles.

Identify: ABCD is cyclic → opposite angles sum to 180° (Theorem 4).

Equation: (4x + 5) + (2x + 35) = 180 → 6x + 40 = 180 → 6x = 140 → x = 23.33…
Hmm — let's use cleaner numbers: if ∠DAB = (3x + 15)° and ∠BCD = (x + 45)°:
(3x + 15) + (x + 45) = 180 → 4x + 60 = 180 → 4x = 120 → x = 30°

Answer: x = 30, ∠DAB = 105°, ∠BCD = 75° [opposite angles in cyclic quadrilateral]

Example 3 — Tangent + Alternate Segment

TAS is a tangent at A. Chord AB subtends angle 40° with the tangent (∠TAS-to-AB angle = 40°). C is on the major arc. Find ∠ACB.

Identify: Tangent–chord angle at A = 40°. C is in the alternate segment.

Apply Theorem 7 (Alternate Segment): The angle between the tangent and the chord AB equals the angle in the alternate segment. → ∠ACB = 40°

Answer: ∠ACB = 40° [alternate segment theorem]

Example 4 — Multi-Step (3 Theorems)

PA and PB are tangents from P to a circle with centre O. ∠APB = 52°. C is on the major arc AB. Find ∠ACB.

Step 1: ∠OAP = ∠OBP = 90° [tangent ⊥ radius at A and B]

Step 2: Quadrilateral OAPB: 52 + 90 + 90 + ∠AOB = 360 → ∠AOB = 128° [angle sum of quadrilateral]

Step 3: ∠ACB = ½ × 128° = 64° [angle at centre = 2 × angle at circumference, arc AB]

Answer: ∠ACB = 64°

Example 5 — Formal Proof Write-Up

Prove that tangents from an external point are equal in length.

Given: Circle centre O, external point P, tangents PA and PB touching circle at A and B.

Step 1: OA ⊥ PA [tangent ⊥ radius, Theorem 5] → ∠OAP = 90°
OB ⊥ PB [tangent ⊥ radius, Theorem 5] → ∠OBP = 90°

Step 2: Consider △OAP and △OBP:
OA = OB [radii of same circle]
OP = OP [common hypotenuse]
∠OAP = ∠OBP = 90° [shown above]

Step 3: By RHS congruence: △OAP ≅ △OBP
Therefore PA = PB (corresponding sides of congruent triangles) ✓

Example 6 — Converse Application

Points A, B, C, D form a quadrilateral where ∠DAB = 115° and ∠BCD = 65°. Show that ABCD is a cyclic quadrilateral.

Check: ∠DAB + ∠BCD = 115° + 65° = 180°

Apply Converse of Theorem 4: If opposite angles of a quadrilateral sum to 180°, then the quadrilateral is cyclic.

Conclusion: Since ∠DAB + ∠BCD = 180°, ABCD is a cyclic quadrilateral (all four vertices lie on a common circle). ✓

Common Mistakes to Avoid

Mistake 1 — Not Stating the Theorem

Writing "∠ACB = 64°" without any reason. The examiner needs to see "angle at centre = 2 × angle at circumference" to award the method mark. Always write the theorem name in square brackets after your answer. This is worth 1 mark per step in most IGCSE mark schemes.

Mistake 2 — Angle at Centre vs. Circumference: Wrong Direction

Students sometimes write: "angle at circumference = 2 × angle at centre" — this is BACKWARDS. The angle at the CENTRE is DOUBLE the angle at the circumference. Centre is bigger. If circumference = 35°, centre = 70°, NOT 17.5°.

Mistake 3 — Alternate Segment: Wrong Side

The alternate segment theorem says the tangent-chord angle equals the angle in the ALTERNATE (opposite) segment. Students often pick the angle in the SAME segment. Draw the chord, identify which side the tangent angle is on, then look at the OTHER side for the equal angle.

Mistake 4 — Cyclic Quadrilateral: Adjacent Not Opposite

Theorem 4 says OPPOSITE angles sum to 180° — not adjacent. In quadrilateral ABCD, the opposite pairs are (A, C) and (B, D). Students sometimes add adjacent angles (A + B) and set them to 180°, which would only be true if the sides were parallel.

Mistake 5 — Missing the Isosceles Triangle

Whenever O (centre) is connected to two points on the circle, those two sides are radii and the triangle is isosceles. The base angles are equal. Not spotting this means missing a step in many multi-step proofs. Always check: "Are any two sides of this triangle both radii?"

Key Formulas — All 7 Circle Theorems

Theorem 1

∠ at centre = 2 × ∠ at circumference

Same arc. O is centre, A and B on circle, C on circle (major arc).

Theorem 2

∠ in semicircle = 90°

AB is diameter. C any point on circle → ∠ACB = 90°.

Theorem 3

∠s in same segment are equal

C, D on same arc, looking at chord AB → ∠ACB = ∠ADB.

Theorem 4

Opposite ∠s in cyclic quad = 180°

ABCD on circle → ∠A + ∠C = 180°; ∠B + ∠D = 180°.

Theorem 5

Tangent ⊥ radius at point of contact

T tangent at P, O centre → OP ⊥ T → ∠OPT = 90°.

Theorem 6

Tangents from external point are equal

P external, tangents to A and B → PA = PB.

Theorem 7 — Alternate Segment Theorem

∠(tangent, chord) = ∠ in alternate segment

The angle between tangent TA and chord AB equals the inscribed angle ACB where C is in the segment on the OTHER side of chord AB from the tangent angle.

Proof techniques summary:
Thm 1: draw radius to C → 2 isosceles △s → exterior angle argument
Thm 2: special case of Thm 1 (AOB = 180°)
Thm 3: both = ½ × same central angle
Thm 4: two applications of Thm 1 → arcs sum to 360°
Thm 5: proof by contradiction
Thm 6: RHS congruence using Thm 5
Thm 7: use diameter + Thm 2 + complementary angles

Circle Theorem Explorer

Theorem: Angle (°): 60°

Select a theorem and move the slider to explore.

Exercise 1 — Each Theorem (Basic)

One question per theorem. Enter your numerical answer.

1. O is the centre. ∠AOB = 80°. Find ∠ACB (C on major arc). [Theorem 1]

2. AB is a diameter. C is on the circle. ∠CAB = 35°. Find ∠CBA. [Theorem 2 — angle in semicircle is 90°]

3. C and D are on the same arc above chord AB. ∠ACB = 47°. Find ∠ADB. [Theorem 3]

4. Cyclic quadrilateral ABCD. ∠DAB = 110°. Find ∠BCD. [Theorem 4]

5. Tangent TP touches circle at P, centre O. ∠TPO = ? degrees. [Theorem 5]

6. PA and PB are tangents from P. PA = 9 cm. Find PB. [Theorem 6]

7. Tangent-chord angle at A = 55°. C in alternate segment. Find ∠ACB. [Theorem 7]

8. O is centre. ∠AOB = 140°. Find ∠ACB where C is on the MINOR arc. [Reflex angle: use 360 − 140 for major, angle = half]

Exercise 2 — Find Angles Using Multiple Theorems

Each question may require more than one theorem. Find the requested angle.

1. O is centre. OA = OB = OC (radii). ∠OAB = 25°. Find ∠AOB. [Isosceles triangle]

2. O is centre. ∠OCA = 30°. ∠OCB = 20°. ∠ACB = ? [Sum of the two base angles]

3. AB is diameter. ∠ABC = 90° [semicircle]. ∠BAC = 40°. Find ∠BCA.

4. Cyclic quad ABCD. ∠ABC = 95°. Find ∠ADC.

5. ∠AOB = 100° (at centre). OA = OB. Find the base angle ∠OAB.

6. C and D on same arc. ∠CAB = 28°. Find ∠DAB if D is also in the same segment (same arc, same chord).

7. PA and PB tangents. ∠OAP = 90°. ∠APB = 40°. ∠AOB = ?

8. Tangent at A. Chord AB. ∠TAB = 38° (tangent-chord). C in same arc (NOT alternate). ∠ACB = 180 − 38 = ? [supplementary in cyclic quad interpretation]

Exercise 3 — Algebra: Solve for x

Each question involves setting up and solving an equation. Enter the value of x.

1. Angle at centre = 4x°. Angle at circumference = (x + 15)°. Theorem 1: 4x = 2(x+15). Solve for x.

2. Cyclic quad: opposite angles (3x + 5)° and (x + 35)°. Sum = 180. Solve for x.

3. Angle at centre = 6x°. Angle at circumference = (2x + 12)°. Solve for x.

4. Cyclic quad: opposite angles (5x − 10)° and (3x + 30)°. Sum = 180. Solve for x.

5. Isosceles triangle (two radii). Base angle = (2x + 5)°. Apex angle at centre = (5x − 10)°. Angles sum to 180. Solve for x.

6. Same segment: ∠ACB = (3x − 4)° and ∠ADB = (x + 20)°. Set equal, solve for x.

7. Tangent-chord = (4x + 3)°. Alternate segment angle = (6x − 11)°. Set equal, solve for x.

8. Centre angle = (10x)°. Circumference angle = (3x + 15)°. Solve for x.

Exercise 4 — Tangent Problems

Problems involving tangents. All lengths in cm unless stated.

1. PA and PB tangents from P. PA = 15 cm. Find PB.

2. Tangent from P to circle, radius = 5. OP = 13. Find tangent length PA (Pythagoras).

3. Radius = 8. OP = 17. Find tangent length PA.

4. PA = PB = 20, OP = 25. Find radius OA.

5. ∠APB = 70°. ∠OAP = 90°. Find ∠AOB (use quadrilateral angle sum).

6. ∠APB = 64°. ∠AOB = ? (use quadrilateral: 64 + 90 + 90 + ∠AOB = 360)

7. ∠AOB = 120°. Find ∠ACB where C is on the major arc [use Theorem 1].

8. Alternate segment: tangent-chord angle = 72°. Find the angle in the alternate segment.

Exercise 5 — Formal Proof Reasoning Chains

Multi-step reasoning. Enter the final numerical answer asked.

1. O is centre. A, B, C on circle. ∠OCA = 22°, ∠OCB = 18°. Find ∠AOB. [Use proof of Theorem 1: ∠AOB = 2(∠OCA + ∠OCB)]

2. AB is diameter. C on circle. ∠ACB = ? degrees (always, by Theorem 2).

3. PA, PB tangents. OA = 6, OP = 10. Find PA (Pythagoras). Then find ∠APB if ∠AOB = 106°.

4. Cyclic quad ABCD. ∠A = 88°, ∠B = 102°. Find ∠C (opposite to A).

5. O is centre. ∠AOC = 160° (reflex is 200°). C on major arc. ∠ABC where B on minor arc = ½ × reflex ∠AOC = ½ × 200 = ?

6. Two tangents from P. ∠APB = 44°. C on circle, major arc. ∠ACB = ? [Find ∠AOB first, then halve]

7. Same segment: ∠PRQ = 53°. Find ∠PSQ (S also on same arc, same chord PQ).

8. ∠OAB = 34° (O centre, A B on circle). Find ∠AOB. [Isosceles: ∠OBA = 34° too]

Practice — 25 Questions

Mixed practice covering all circle theorem skills.

1. ∠AOB = 90° (centre). Find ∠ACB (circumference).

2. ∠ACB = 35° (circumference). Find ∠AOB (centre).

3. AB is diameter. ∠CAB = 55°. Find ∠CBA (angle sum in triangle, ∠ACB = 90°).

4. Same segment: ∠ACB = 61°. Find ∠ADB.

5. Cyclic quad: ∠A = 75°. Find ∠C.

6. Cyclic quad: ∠B = 88°. Find ∠D.

7. Tangent-chord angle = 48°. Alternate segment angle = ?

8. PA = PB (tangents). PA = 11. Find PB.

9. OA = 5, OP = 13 (external). Find tangent length PA.

10. ∠AOB = 74° (centre angle). Find ∠ACB (circumference).

11. ∠ACB = 53° (circumference). Find ∠AOB (centre).

12. Isosceles OAB: ∠OAB = 40°. Find ∠AOB.

13. ∠APB = 50° (tangents from P). Find ∠AOB.

14. ∠AOB = 130°. C on major arc. Find ∠ACB.

15. AB diameter. ∠BAC = 28°. Find ∠ABC.

16. Tangent-chord angle = 33°. Find angle in alternate segment.

17. Cyclic quad: ∠A = 104°, ∠B = 76°. Find ∠C.

18. Same arc. ∠ACB = 29°. Find ∠ADB.

19. ∠OCA = 15°, ∠OCB = 25°. Find ∠AOB using Theorem 1 proof.

20. ∠AOB (reflex) = 260°. Find ∠ACB (C on minor arc side) = ½ × 260.

21. Solve for x: angle at centre = 8x, at circumference = (3x + 5). Equation: 8x = 2(3x+5). Value of x?

22. Cyclic quad: opposite angles (4x + 10) and (2x + 20). Sum = 180. Find x.

23. PA tangent, OA = 9, OP = 15. Find PA.

24. ∠APB = 80°. Find ∠AOB (2 tangents from P, quadrilateral method).

25. ∠OAB = 52°. OA = OB. Find ∠AOB.

Challenge — 12 Questions

Harder multi-step and algebraic problems. Requires careful reasoning.

1. O is centre. ∠OAB = 38°, ∠OBC = 24°. A, B, C on circle. Find ∠AOC (use two isosceles triangles, then angles at O).

2. PA and PB tangents. ∠OAB = 32°. Find ∠APB. [Hint: OA = OB → isosceles; ∠AOB = 180−64 = 116; ∠APB = 360−90−90−116]

3. Cyclic quad ABCD. ∠A = (5x − 5)°, ∠C = (3x + 25)°. Solve for x.

4. Centre angle = 5x. Circumference = (x + 40)°. Solve for x.

5. O is centre. A, B on circle. OA = OB. ∠AOB = 108°. Find ∠ACB where C is on the major arc.

6. Tangent-chord at A = (3x + 7)°. Alternate segment angle = (5x − 13)°. Solve for x.

7. Two chords AC and BD intersect inside circle at E. ∠AEB = 75°. ∠CAD = 40°. Find ∠ABD.

8. OA = 7, OP = 25. Two tangents from P. Find PA. Then find ∠OAP (it's 90 — enter 90).

9. Cyclic quad: ∠A + ∠C = 180. ∠A = 3∠C. Find ∠A.

10. ∠ACB = (x + 10)°. ∠ADB = (2x − 15)° (same segment). Solve for x.

11. ∠APB = 36°. C on circle, major arc. ∠AOB = 360 − 90 − 90 − 36 = 144. ∠ACB = ?

12. AB diameter = 20 cm. AC = 12 cm. ∠ACB = 90°. Find BC (Pythagoras).

Exam Style — 5 Questions

IGCSE-style questions requiring full reasoning. Enter the numerical answer.

[5 marks] O is the centre of a circle. A, B, C are on the circle. OA = OC. ∠OCA = 28°. Find ∠ABC.
Hint: isosceles → ∠OAC = 28°; ∠AOC = 124°; ∠ABC = ½ × ∠AOC

[4 marks] ABCD is a cyclic quadrilateral. ∠DAB = (4x + 12)°, ∠BCD = (2x + 24)°. Find ∠DAB.
Hint: sum = 180; solve for x; substitute back

[4 marks] TP is a tangent to circle centre O at P. OP = 8 cm, OT = 17 cm. Find TP.

[5 marks] Tangent at A. ∠TAB = 58° (tangent–chord). B and C are on circle. ∠ABC = 64°. Find ∠ACB.
Hint: ∠CAB = ∠TAB = 58° (alt. segment); angle sum in △ABC: ∠ACB = 180 − 58 − 64

[6 marks] PA and PB are tangents from external point P. ∠APB = 48°. C is on the major arc. Find ∠ACB.
Hint: quadrilateral OAPB; ∠AOB = 360 − 90 − 90 − 48 = 132; ∠ACB = 66