Chapter 7 of 10

🔵 Circles & Trigonometry

7 min+20 XP

This chapter covers circle properties, angle properties of circles, Pythagoras' theorem, trigonometric ratios (SOH CAH TOA), sine rule, cosine rule, area of triangle, and bearings. These topics carry very heavy marks in Paper 2 and require both conceptual understanding and computational accuracy.

Circle theorems are a favourite topic for examiners because they test geometric reasoning. Each theorem must be quoted precisely in your working. Trigonometry, on the other hand, is a computational skill — you need to know when to use SOH CAH TOA (right-angled triangles only) versus the sine rule or cosine rule (any triangle).

Circle Theorems

Circle theorems describe the relationships between angles, chords, and tangents in a circle. There are six main theorems you must know. In the exam, you must state the theorem name and apply it correctly. Many questions combine two or three theorems in a single problem.

OABCθ∠AOB = 2 × ∠ACB (angle at centre theorem)

The angle at the centre is twice the angle at the circumference subtended by the same arc

6 Essential Circle Theorems

1
Angle at centre = 2 x angle at circumference

The angle subtended by an arc at the centre of the circle is exactly twice the angle subtended at any point on the circumference, from the same arc.

2
Angle in a semicircle = 90 degrees

Any angle inscribed in a semicircle (i.e., subtended by a diameter at the circumference) is always a right angle.

3
Angles in same segment are equal

All angles subtended by the same arc (or chord) at points on the same side of the chord are equal.

4
Opposite angles of cyclic quadrilateral sum to 180 degrees

A cyclic quadrilateral has all four vertices on the circle. Its opposite angles are supplementary.

5
Tangent is perpendicular to radius

A tangent to a circle is perpendicular (90 degrees) to the radius drawn to the point of tangency.

6
Tangents from external point are equal

Two tangent lines drawn from the same external point to a circle have equal length.

Circle Theorems

Essential circle properties tested in O-Level E-Math. Must know all theorems and their converses.

Angle at centre = 2 x angle at circumference
Angle in a semicircle
Angles in the same segment
Opposite angles of a cyclic quadrilateral
Tangent perpendicular to radius
Two tangents from an external point
Alternate segment theorem

📝Worked Example 1: Angle at Centre Theorem

Q
Question

In the circle with centre O, angle AOB = 124 degrees. C is a point on the major arc. Find angle ACB.

124°62°OABCCentral angle = 124°Inscribed angle = 62°

Angle at centre is twice the angle at the circumference

📝Worked Example 2: Cyclic Quadrilateral

Q
Question

ABCD is a cyclic quadrilateral. Angle A = 110 degrees and angle B = 85 degrees. Find angles C and D.

110°85°70°95°ABCDA + C = 180° | B + D = 180°

Opposite angles of a cyclic quadrilateral sum to 180°

📝Worked Example 3: Tangent-Radius Problem

Q
Question

A tangent touches a circle with centre O at point T. OT = 5 cm and the tangent meets a line from O at point P, where OP = 13 cm. Find PT.

5 cm12 cm13 cmOTPtangentOT ⊥ tangent | PT² = OP² − OT²

Tangent is perpendicular to radius at the point of contact

📝Worked Example 3b: Angles in the Same Segment

Q
Question

Points A, B, C, and D lie on a circle. AC and BD intersect at E inside the circle. Angle AEB = 70 degrees and angle ABD = 40 degrees. Find angle ACD.

OABCθ∠AOB = 2 × ∠ACB (angle at centre theorem)

The angle at the centre is twice the angle at the circumference subtended by the same arc

📝Worked Example 3c: Alternate Segment Theorem

Q
Question

A tangent at point T makes angle 55 degrees with chord TA. B is a point on the circle on the opposite side of chord TA from the tangent. Find angle ABT.

55°55°TABtangentAngle between tangent & chord = angle in alternate segment

Alternate segment theorem: both angles equal 55°

📝Worked Example 3d: Multi-Theorem Circle Problem

Q
Question

ABCD is a cyclic quadrilateral. The tangent at A makes angle 50 degrees with AB. Angle ADC = 115 degrees. Find (a) angle ACB, (b) angle ABC.

110°85°70°95°ABCDA + C = 180° | B + D = 180°

Opposite angles of a cyclic quadrilateral sum to 180°

Circle Theorem Exam Language

Examiners award marks for precise theorem names. Use these exact phrasings:

  • "Angle at centre = 2 x angle at circumference"
  • "Angle in a semicircle = 90 degrees"
  • "Angles in the same segment are equal"
  • "Opposite angles of a cyclic quadrilateral are supplementary"
  • "Tangent perpendicular to radius"
  • "Tangents from an external point are equal in length"
  • "Alternate segment theorem"

Do NOT use vague language like "the angles are the same" or "they add up to 180". Be specific about WHICH theorem you are using.

🧠Quick Check

A tangent at point P makes angle 65 degrees with chord PQ. R is on the circle in the alternate segment. What is angle PRQ?

A65 degrees
B115 degrees
C130 degrees
D25 degrees

Pythagoras' Theorem & Trigonometric Ratios

Pythagoras' theorem states that in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. This is the foundation for all distance calculations in coordinate geometry and many real-world problems. Trigonometric ratios (SOH CAH TOA) relate the angles and sides of right-angled triangles.

📏Pythagoras Theorem

Relates the sides of a right-angled triangle. Foundation for trigonometry and coordinate geometry.

Pythagoras' theorem
Finding a shorter side

📊Trigonometric Ratios (SOH CAH TOA)

For right-angled triangles. The three primary trigonometric ratios.

Sine
Cosine
Tangent
Relationship

📝Worked Example 4: SOH CAH TOA

Q
Question

In right triangle PQR, angle Q = 90 degrees, PQ = 8 cm, and angle P = 35 degrees. Find QR and PR.

35°PQRPQ = 8 cmQR = ?PR = ?adjacentoppositehypotenuse

Right triangle PQR: identify opposite, adjacent and hypotenuse relative to angle P

📝Worked Example 5: Pythagoras in 3D

Q
Question

A rectangular box has length 12 cm, width 4 cm, and height 3 cm. Find the length of the space diagonal (from one corner to the opposite corner through the interior).

ABCDEFGH8 cm6 cm5 cmBase diagonal ACSpace diagonal AG

Cuboid with base diagonal AC and space diagonal AG

📝Worked Example 5b: Angle of Elevation

Q
Question

A boy stands 50 m from the base of a building. The angle of elevation to the top is 32 degrees. Find the height of the building.

32°50 mh = ?Angle of elevation = 32°

Angle of elevation from a point on the ground to the top of a building

📝Worked Example 5c: Angle of Depression

Q
Question

From the top of a cliff 80 m high, the angle of depression to a boat at sea is 25 degrees. How far is the boat from the base of the cliff?

32°50 mh = ?Angle of elevation = 32°

Angle of elevation from a point on the ground to the top of a building

📝Worked Example 5d: Finding an Angle Using Inverse Trig

Q
Question

A ladder 5 m long leans against a wall. The foot of the ladder is 3 m from the wall. Find (a) how high up the wall the ladder reaches, and (b) the angle the ladder makes with the ground.

35°PQRPQ = 8 cmQR = ?PR = ?adjacentoppositehypotenuse

Right triangle PQR: identify opposite, adjacent and hypotenuse relative to angle P

Exact Trigonometric Values

You must memorise the exact values for the standard angles: 0 degrees, 30 degrees, 45 degrees, 60 degrees, and 90 degrees. These come up frequently in both Paper 1 (non-calculator) and Paper 2. The examiners expect exact values (surds, not decimals) when these special angles appear.

Anglesincostan
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Memory Tricks for Exact Values

For sine of 0°, 30°, 45°, 60°, 90°: the numerators follow the pattern , all divided by 2. This gives . For cosine, it is the same sequence but reversed. For tan, simply divide sin by cos for each angle.

SOH CAH TOA — How to Remember

Some Old Horse Caught Another Horse Taking Oats Away

Or simply: SOH-CAH-TOA (say it like a word: "so-ka-toa")

  • SOH:
  • CAH:
  • TOA:

Exam trick: To decide which ratio to use, identify which two of the three sides (Opposite, Adjacent, Hypotenuse) are involved in your question. The ratio that uses those two sides is the one you need.

🧠Quick Check

What is as an exact value?

A
B
C
D
0°90°180°270°360°1-1y = sin xy = cos xPeriod = 360°

Sine (blue) and cosine (red) curves with period 360° and amplitude 1

Sine Rule & Cosine Rule

SOH CAH TOA only works for right-angled triangles. For any other triangle, you need the sine rule or cosine rule. The choice depends on what information you have.

When to Use Sine Rule vs Cosine Rule

Use Sine Rule when you have a matched pair (a side and its opposite angle) plus one more piece. Use Cosine Rule when you have SAS (two sides and the included angle) or SSS (all three sides). If the triangle has a right angle, just use SOH CAH TOA — it is simpler.

🔺Sine Rule & Cosine Rule (Non-Right Triangles)

For any triangle (not just right-angled). Used when SOH CAH TOA cannot be applied directly.

Sine rule
Sine rule (finding angles)
Cosine rule (finding side)
Cosine rule (finding angle)
Area of triangle (using sine)

📝Worked Example 6: Cosine Rule (Finding a Side)

Q
Question

In triangle ABC, b = 8 cm, c = 5 cm, angle A = 60 degrees. Find side a.

📝Worked Example 7: Cosine Rule (Finding an Angle)

Q
Question

In triangle PQR, p = 7, q = 8, r = 13. Find angle R.

📝Worked Example 8: Sine Rule (Finding an Angle)

Q
Question

In triangle PQR, p = 10, q = 7, angle P = 50 degrees. Find angle Q.

📝Worked Example 9: Area Using Sine

Q
Question

Find the area of triangle ABC where a = 12, b = 9, and angle C = 40 degrees.

📝Worked Example 9b: Sine Rule (Finding a Side)

Q
Question

In triangle ABC, angle A = 42 degrees, angle B = 73 degrees, and side a = 15 cm. Find side b.

📝Worked Example 9c: Real-World Cosine Rule Application

Q
Question

Two ships leave port P at the same time. Ship A travels on bearing 060 degrees at 15 km/h for 2 hours. Ship B travels on bearing 150 degrees at 10 km/h for 2 hours. Find the distance between the ships after 2 hours.

NN060°150°30 km20 kmPABPAB = 90°

Bearing: P to A (060°), A to B (150°)

The Ambiguous Case of Sine Rule

When using the sine rule to find an angle, there may be two possible answers: one acute and one obtuse. This happens because sin x = sin(180 - x). You must check whether the obtuse angle is valid by adding it to the known angles and checking if the sum is less than 180 degrees. If both are valid, state both solutions. In O-Level exams, the question usually specifies "acute" or "obtuse", or the diagram makes it clear which one to use.

Decision Flowchart: Which Formula to Use?

1
Is the triangle right-angled?

YES: Use SOH CAH TOA (for sides/angles) or Pythagoras (for sides only). These are simpler and should be preferred.

2
Do you have a matched pair (side + opposite angle)?

YES: Use Sine Rule. You need at least one matched pair plus one extra piece of information.

3
Do you have SAS (two sides + included angle)?

YES: Use Cosine Rule to find the third side. Or use the area formula (1/2 ab sin C) for the area.

4
Do you have SSS (all three sides)?

YES: Use Cosine Rule to find any angle.

5
Do you need the area?

Use Area = 1/2 ab sin C if you have SAS. Otherwise, use Area = 1/2 base x height if you know the base and height.

🧠Quick Check

In triangle ABC, a = 10, angle A = 30 degrees, angle B = 45 degrees. Which formula should you use to find side b?

ASOH CAH TOA
BCosine rule
CSine rule
DPythagoras

Bearings

Bearings are measured clockwise from North and always written as 3-digit numbers (e.g., 045 degrees, 130 degrees, 270 degrees). The bearing of B from A means: stand at A, face North, then measure the angle clockwise to the direction of B. To find the back bearing (bearing of A from B), add or subtract 180 degrees.

🧭Bearings & Angles of Elevation/Depression

Bearings are measured clockwise from North. Angles of elevation/depression are measured from the horizontal.

Bearing convention
Back bearing
Angle of elevation
Angle of depression

📝Worked Example 10: Bearing Problem

Q
Question

A ship sails from P on bearing 040 degrees for 80 km to Q. It then sails on bearing 130 degrees for 60 km to R. Find PR and the bearing of R from P.

NN040°130°80 km60 kmPQRPQR = 90°

Bearing: P to Q (040°), Q to R (130°)

📝Worked Example 10b: Three-Point Bearing with Cosine Rule

Q
Question

From point A, point B is on bearing 060 degrees at distance 8 km. From A, point C is on bearing 150 degrees at distance 10 km. Find (a) angle BAC and (b) distance BC.

NN060°150°8 km10 kmABCABC = 90°

Bearing: A to B (060°), B to C (150°)

📝Worked Example 10c: Non-Right Bearing Triangle

Q
Question

A hiker walks from P to Q on bearing 075 degrees for 6 km, then from Q to R on bearing 200 degrees for 4 km. Find the distance PR and the bearing of R from P.

NN075°200°6 km4 kmPQRPQR = 55°

Bearing: P to Q (075°), Q to R (200°)

Bearing Problem Checklist

Follow this checklist for every bearing question:

  1. Draw a clear diagram with North lines at EVERY point
  2. Mark all given bearings and distances
  3. Find angles inside the triangle using back bearings (add or subtract 180 degrees)
  4. Use the appropriate formula: Pythagoras, sine rule, cosine rule
  5. Express the final bearing as a 3-digit number (e.g., 076 degrees, not 76 degrees)
🧠Quick Check

The bearing of B from A is 310 degrees. What is the bearing of A from B?

A130 degrees
B310 degrees
C490 degrees
D050 degrees

⚠️Common Mistakes — Circles & Trig

Common Mistake
Using sine rule when you have SAS (two sides + included angle).
Correct
Use COSINE rule for SAS. Sine rule is for AAS or ASA.

Why: Sine rule: need a side-angle pair. Cosine rule: need SAS or SSS. Using the wrong rule gives incorrect answers.

Common Mistake
Bearing of , written as .
Correct
Bearing must be written as (always 3 digits).

Why: Bearings are ALWAYS written as 3-digit numbers, measured clockwise from North. Use leading zeros: 045, 090, 005.

Common Mistake
Area of circle
Correct
Area . Circumference .

Why: Area uses r-squared, circumference uses 2r. Remember: area = pi r^2, circumference = 2 pi r.

Common Mistake
Volume of cone
Correct
Volume of cone

Why: A cone is one-third of a cylinder with the same base and height. Forgetting the 1/3 is a frequent error.

Common Mistake
Surface area of sphere
Correct
Surface area of sphere

Why: The coefficient is 4, not 2. Hemisphere curved SA = 2 pi r^2, but a full sphere is 4 pi r^2.

3D Trigonometry Problems

Three-dimensional trigonometry questions involve finding lengths or angles in solid shapes such as cuboids, pyramids, and prisms. The technique is always the same: identify right-angled triangles within the 3D shape and apply Pythagoras or SOH CAH TOA to each triangle separately. You usually need to find an intermediate length (such as a diagonal of a face) before you can find the required answer.

📝Worked Example 11: Angle in a Cuboid

Q
Question

A cuboid has dimensions 6 cm by 8 cm by 5 cm. Find the angle that the space diagonal AG makes with the base ABCD.

ABCDEFGH8 cm6 cm5 cmBase diagonal ACSpace diagonal AG

Cuboid with base diagonal AC and space diagonal AG

VABCDOM12 cm5 cm13 cm10 cmVO base, VM² = VO² + OM² (Pythagoras)

Square-based pyramid with VO = 12 cm, OM = 5 cm, VM = 13 cm

📝Worked Example 12: Angle in a Pyramid

Q
Question

A square-based pyramid has base side 10 cm and vertical height 12 cm. Find (a) the slant height and (b) the angle between a triangular face and the base.

Strategy for 3D Trigonometry

1
Step 1: Draw a clear 3D diagram

Label all given measurements. Use dotted lines for hidden edges.

2
Step 2: Identify the right-angled triangles

Most 3D trig problems reduce to 2 or 3 right-angled triangles. The triangles are usually on the faces or cross-sections of the solid.

3
Step 3: Find intermediate lengths first

You often need to find a base diagonal or face diagonal before you can find the final angle or length.

4
Step 4: Apply Pythagoras or SOH CAH TOA

Work through each right triangle separately. State which triangle you are working with.

5
Step 5: State the required angle clearly

Describe the angle using three letters (e.g., angle GAC) or a clear verbal description.

🧠Quick Check

A cuboid is 3 cm by 4 cm by 12 cm. Find the length of the space diagonal.

A13 cm
B19 cm
C5 cm
D7 cm

Cosine Rule: Finding an Angle

The cosine rule can be rearranged to find an angle when all three sides are known. This is a very common Paper 2 question. Remember to use the rearranged form: . The side opposite the angle you are finding goes in the numerator with a MINUS sign.

📝Worked Example 13: Cosine Rule for Angle

Q
Question

In triangle PQR, PQ = 7 cm, QR = 9 cm, PR = 11 cm. Find angle PQR.

📝Worked Example 14: Area of Triangle Using Sine

Q
Question

In triangle ABC, AB = 8 cm, AC = 12 cm, angle BAC = 53 degrees. Find the area of the triangle.

🧠Quick Check

In triangle ABC, all three sides are known but no angles. To find angle B, which formula do you use?

A
BSine rule:
CCosine rule:
DArea
🎯Key Takeaway
Memorise all 6 circle theorems and quote them precisely. Use SOH CAH TOA for right-angled triangles, sine rule for matched pairs, and cosine rule for SAS or SSS. For bearings, always measure clockwise from North and write as 3 digits. For 3D trig, identify right-angled triangles within the solid and find intermediate lengths first. For area of a triangle with SAS, use (1/2)ab sin C. Check your calculator is in degree mode!