← IGCSE / O Level Mathematics

IGCSE / O Level Mathematics · Topic 6 Trigonometry · Chapter 29 · pp. 394–417

Trigonometry

Trigonometry links the angles of a right-angled triangle to the lengths of its sides. Name the three sides from the angle you care about, pick sine, cosine or tangent, and one short equation gives you the missing side or angle. Pythagoras' theorem handles the cases with no angle at all.

Learning objectives

  1. Name the hypotenuse, opposite and adjacent sides of a right-angled triangle for a chosen angle.
  2. Use tangent, sine and cosine to find a missing side or a missing angle.
  3. Use Pythagoras' theorem, and decide when to use it instead of trigonometry.
  4. Solve problems with bearings and with angles of elevation and depression.
  5. Sketch the sine, cosine and tangent curves for 0° to 360° and solve trigonometric equations Extended.

Book exercises (pp. 394–417): 29.1 to 29.9 and Student assessments 1, 2 and 3. The curves and equations (29.7 to 29.9) and Student assessment 3 are Extended content.

💡 Give answers to 3 s.f. and angles to 1 d.p. unless told otherwise. Check your calculator is in degree mode (it shows D or DEG).

Naming the sides

🔑 Key ideas

  • The hypotenuse is the longest side, always opposite the right angle.
  • The opposite side is across from the angle you chose.
  • The adjacent side is next to the chosen angle (and is not the hypotenuse).
  • Choose a different angle and opposite and adjacent swap. The hypotenuse never changes.
θ hypotenuse opposite adjacent θ hypotenuse opposite adjacent
The same triangle twice. On the left θ is at the bottom-left corner. On the right θ is at the top corner, so opposite and adjacent have swapped.
RatioFormulaMemoryUse it when you have
sinesin θ = SOHO and H
cosinecos θ = CAHA and H
tangenttan θ = TOAO and A

The tangent ratio

📘 Example 1 · Find the angle: opposite 4 cm, adjacent 5 cm

  1. The sides are opposite and adjacent, so use tangent (TOA).
  2. tan x° = .
  3. x = tan−1() = 38.66…
  4. Angle BAC = 38.7° (1 d.p.).

📘 Example 2 · Find the angle: opposite 8 cm, adjacent 3 cm

  1. tan x° = .
  2. x = tan−1() = 69.44…
  3. Angle BAC = 69.4° (1 d.p.).

📘 Example 3 · Find the opposite side QR: angle 42°, adjacent 6 cm

  1. Opposite and adjacent: use tangent. tan 42° = .
  2. The unknown is on top, so multiply both sides by 6: p = 6 × tan 42°.
  3. p = 6 × 0.9004 = 5.402…
  4. QR = 5.40 cm (3 s.f.).

📘 Example 4 · Find the adjacent side XY: angle 35°, opposite 6 cm

  1. tan 35° = . The unknown is on the bottom.
  2. Multiply both sides by z: z × tan 35° = 6.
  3. Divide both sides by tan 35°: z = .
  4. z = 6 ÷ 0.7002 = 8.569…
  5. XY = 8.57 cm (3 s.f.).

✏️ Try it

  1. tan 50° = . Find x to 3 s.f.
  2. A right-angled triangle has opposite side 9 cm and adjacent side 4 cm. Find the angle, to 1 d.p.
  3. The angle is 25° and the opposite side is 7 cm. Find the adjacent side, to 3 s.f.
Answers

1. x = 10 × tan 50° = 10 × 1.1918 = 11.9 · 2. tan⁻¹(9 ÷ 4) = tan⁻¹(2.25) = 66.0° · 3. 7 ÷ tan 25° = 7 ÷ 0.4663 = 15.0

Next: sine and cosine →

Sine and cosine

🔑 Key ideas

  • Sine uses the opposite side and the hypotenuse: sin θ = O ÷ H.
  • Cosine uses the adjacent side and the hypotenuse: cos θ = A ÷ H.
  • Finding a side: write the ratio, rearrange, calculate.
  • Finding an angle: write the ratio as a fraction, then press sin−1, cos−1 or tan−1.
Finding a side Label O, A, H Pick the ratio Put in the numbers Rearrange and calculate Finding an angle Label O, A, H Pick the ratio Ratio = a fraction Press sin⁻¹, cos⁻¹ or tan⁻¹
Both routes start the same way: label the sides, then pick the ratio from the two sides that are involved.

📘 Example 5 · Find angle BAC: opposite 7 cm, hypotenuse 12 cm

  1. Opposite and hypotenuse: use sine (SOH).
  2. sin x = .
  3. x = sin−1() = 35.68…
  4. Angle BAC = 35.7° (1 d.p.).

📘 Example 6 · Find the hypotenuse PR: angle 18°, opposite 11 cm

  1. Opposite and hypotenuse: sin 18° = . The unknown is on the bottom.
  2. Multiply by q: q × sin 18° = 11.
  3. Divide by sin 18°: q = = 11 ÷ 0.3090 = 35.60…
  4. PR = 35.6 cm (3 s.f.).

📘 Example 7 · Find XY: angle 62°, hypotenuse 20 cm, XY adjacent

  1. Adjacent and hypotenuse: cos 62° = .
  2. Multiply by 20: z = 20 × cos 62° = 20 × 0.4695 = 9.390…
  3. XY = 9.39 cm (3 s.f.).

📘 Example 8 · Find angle ABC: adjacent 5.3 m, hypotenuse 12 m

  1. cos x = = 0.4417…
  2. x = cos−1() = 63.79…
  3. Angle ABC = 63.8° (1 d.p.).

📘 Example 9 · Your turn to choose: angle 35°, hypotenuse 9 cm. Find the adjacent side

  1. Adjacent and hypotenuse: cosine.
  2. cos 35° = , so A = 9 × cos 35°.
  3. A = 9 × 0.8192 = 7.372…
  4. Adjacent = 7.37 cm (3 s.f.). Check: it is shorter than the hypotenuse (9 cm). ✔

💡 Sense check: a short side must be shorter than the hypotenuse. If your answer for a leg is bigger than the hypotenuse, you picked the wrong ratio or divided the wrong way.

✏️ Try it

  1. The hypotenuse is 20 cm and an angle is 40°. Find the opposite side, to 3 s.f.
  2. The adjacent side is 6 cm and the hypotenuse is 10 cm. Find the angle, to 1 d.p.
  3. The opposite side is 8 cm and the angle is 52°. Find the hypotenuse, to 3 s.f.
Answers

1. 20 × sin 40° = 20 × 0.6428 = 12.9 · 2. cos⁻¹(6 ÷ 10) = cos⁻¹(0.6) = 53.1° · 3. 8 ÷ sin 52° = 8 ÷ 0.7880 = 10.2

Next: Pythagoras and the solver →

Pythagoras' theorem

🔑 Key ideas

  • In a right-angled triangle: a² + b² = c², where c is the hypotenuse.
  • Finding the hypotenuse: add the squares, then square root.
  • Finding a shorter side: subtract the squares, then square root.
  • Pythagoras uses two sides only. When an angle is involved, use trigonometry.
Right-angled triangle Is there an angle other than 90° in the question? Do you know two sides and want the third side? Yes: sin, cos or tan Yes: Pythagoras
Two sides and no angle: Pythagoras. A side and an angle, or you want an angle: trigonometry.

📘 Example 10 · Find BC: the shorter sides are 8 m and 6 m

  1. BC is the longest side, so it is the hypotenuse: a² = 8² + 6².
  2. a² = 64 + 36 = 100.
  3. a = √100 = 10 m.

📘 Example 11 · Find AC: hypotenuse 12 m, other side 5 m

  1. The hypotenuse is given, so subtract: b² = 12² − 5².
  2. b² = 144 − 25 = 119.
  3. b = √119 = 10.90…
  4. AC = 10.9 m (3 s.f.).

📘 Example 12 · Village A is 30 km due north of C and B is 65 km due east of A. Find CB

  1. North and east meet at a right angle at A, so triangle ACB has a right angle at A.
  2. CB is the hypotenuse: CB² = 30² + 65² = 900 + 4225 = 5125.
  3. CB = √5125 = 71.58…
  4. CB = 71.6 km (nearest 0.1 km).

📘 Example 13 · Mixed: legs 6 cm and 8 cm. Find the hypotenuse and the smallest angle

  1. Hypotenuse: √(6² + 8²) = √100 = 10 cm.
  2. The smallest angle is opposite the shortest side (6 cm), so use sine: sin θ = .
  3. θ = sin−1(0.6) = 36.9° (1 d.p.).
  4. Check with tangent: tan⁻¹(6 ÷ 8) = tan⁻¹(0.75) = 36.9°. ✔

✏️ Try it

  1. The shorter sides are 5 cm and 12 cm. Find the hypotenuse.
  2. The hypotenuse is 25 cm and one shorter side is 7 cm. Find the other side.
  3. The two shorter sides are both 1 cm. Find the hypotenuse, exactly and to 3 s.f.
Answers

1. √(25 + 144) = √169 = 13 cm · 2. √(625 − 49) = √576 = 24 cm · 3. √2 = 1.41 cm

Try it: right-angled triangle solver

🧪 Fill in exactly two boxes and leave the others empty

θ O A H
The solved triangle: O is opposite θ, A is adjacent to θ, H is the hypotenuse. Not drawn to scale.

Working

    ✏️ Try it with the solver

    1. Enter θ = 30 and H = 10. Which ratio does the solver choose for O? Check by hand: 10 × sin 30°.
    2. Enter O = 5 and A = 12. Why does it use Pythagoras first?
    3. Enter O = 9 and H = 6. What does the solver say, and why?
    Answers

    1. Sine (SOH): O = 5. · 2. It needs H before it can find the angle, and it has two sides and no angle: H = 13, θ = 22.6°. · 3. The hypotenuse is the longest side, so O must be smaller than H; the triangle cannot exist.

    Next: elevation and depression →

    Angles of elevation and depression

    🔑 Key ideas

    • The angle of elevation is measured up from the horizontal to the line of sight.
    • The angle of depression is measured down from the horizontal.
    • Both angles are measured from a horizontal line, never from the vertical.
    • The depression from A to B equals the elevation from B to A (alternate angles).
    • Sketch the vertical right-angled triangle first, then label O, A, H.
    θ 60 m (adjacent) h (opposite) line of sight X top
    Angle of elevation of the top of a tower from X. The tower is vertical, the ground horizontal: h = 60 × tan 40°.
    θ θ plane P X on the ground depression elevation
    The two horizontals are parallel, so the angle of depression of X from P equals the angle of elevation of P from X.

    📘 Example 14 · A tower is 60 m from X. The angle of elevation of its top is 40°. Find its height

    1. Sketch the vertical triangle: adjacent = 60 m, opposite = h, angle 40°.
    2. Opposite and adjacent: tan 40° = .
    3. h = 60 × tan 40° = 60 × 0.8391 = 50.35…
    4. The tower is 50.3 m high (3 s.f.).

    📘 Example 15 · A plane sees X at an angle of depression of 30°. The line of sight is 6 km long. Find the plane's height

    1. The depression is measured from the horizontal at the plane, so it equals the 30° angle at X in the triangle.
    2. Height h is opposite, 6 km is the hypotenuse: sin 30° = .
    3. h = 6 × sin 30° = 6 × 0.5 = 3 km.

    📘 Example 16 · Villages A and B are 12 km apart horizontally and 2 km apart vertically. Find the distance AB and the angle of elevation of B from A

    1. Two sides are known and the third is wanted: Pythagoras. AB² = 12² + 2² = 144 + 4 = 148.
    2. AB = √148 = 12.2 km (3 s.f.).
    3. For the angle, opposite = 2 and adjacent = 12: tan θ = .
    4. θ = tan−1(0.1667) = 9.5° (1 d.p.).

    📘 Example 17 · Two planes above P at elevations 65° and 75°, 2 km away horizontally. Find the vertical distance between them

    1. Each plane gives its own right-angled triangle with adjacent = 2 km.
    2. Higher plane: 2 × tan 75° = 2 × 3.7321 = 7.464 km.
    3. Lower plane: 2 × tan 65° = 2 × 2.1445 = 4.289 km.
    4. Subtract: 7.464 − 4.289 = 3.175, so about 3.17 km (3 s.f.).

    ✏️ Try it

    1. A boat is 100 m from the foot of a 40 m cliff. Find the angle of depression of the boat from the top, to 1 d.p.
    2. You stand 20 m from a flagpole and the angle of elevation of its top is 35°. Find the height, to 3 s.f.
    3. A kite string is 50 m long and makes 60° with the ground. How high is the kite?
    Answers

    1. tan θ = 40 ÷ 100 = 0.4, θ = 21.8° · 2. 20 × tan 35° = 20 × 0.7002 = 14.0 m · 3. 50 × sin 60° = 50 × 0.8660 = 43.3 m

    Next: the curves →

    The sine, cosine and tangent curves Extended

    🔑 Key ideas

    • A calculator gives only one angle. The graph shows there are usually two between 0° and 360°.
    • y = sin x: period 360°, maximum 1 at 90°, minimum −1 at 270°.
    • y = cos x: period 360°, maximum 1, minimum −1.
    • y = tan x: period 180°, no maximum or minimum, undefined at 90° and 270°.
    • Symmetry rules: sin x = sin (180° − x) and cos x = −cos (180° − x).
    0° 90° 180° 270° 360° 1 −1 30° 150° 0.5 y = sin x sin 30° = sin 150° = 0.5
    Sine curve. A horizontal line at 0.5 cuts it at 30° and 150°, which add up to 180°.
    0° 90° 180° 270° 360° 1 −1 60° 120° 0.5 −0.5 y = cos x cos 60° = 0.5 and cos 120° = −0.5, so cos 60° = −cos 120°
    Cosine curve. It starts at 1, crosses zero at 90° and is negative between 90° and 270°.
    0° 90° 180° 270° 360° asymptote x = 90° asymptote x = 270° y = tan x
    Tangent curve: it repeats every 180° and never touches the dashed asymptotes.

    📘 Example 18 · sin 30° = 0.5. Which other angle from 0° to 180° has sine 0.5?

    1. Sine is symmetric about 90°: sin x = sin (180° − x).
    2. 180° − 30° = 150°.
    3. Read it off the graph: the line y = 0.5 meets the curve again at 150°.
    4. So sin 150° = 0.5.

    📘 Example 19 · cos 60° = 0.5. Which angle from 0° to 180° has cosine −0.5?

    1. Cosine uses a sign flip: cos x = −cos (180° − x).
    2. 180° − 60° = 120°, so cos 120° = −cos 60° = −0.5.
    3. The angle is 120°, and cos 60° = −cos 120°.

    📘 Example 20 · Express −cos 50° as the cosine of another angle between 0° and 180°

    1. Use cos (180° − x) = −cos x.
    2. 180° − 50° = 130°.
    3. cos 130° = −cos 50°.

    📘 Example 21 · Why is tan 90° undefined?

    1. tan θ = opposite ÷ adjacent.
    2. As θ grows towards 90°, the adjacent side shrinks towards 0 and the triangle becomes a vertical line.
    3. At 90° you would divide by 0, which has no answer.
    4. So tan 90° is undefined: the graph has an asymptote there.

    ✏️ Try it

    1. Write sin 20° as the sine of another angle between 0° and 180°.
    2. Write cos 75° as minus the cosine of another angle.
    3. sin θ = 0.8 for θ between 0° and 180°. Give both angles to the nearest degree.
    Answers

    1. sin 160° · 2. cos 75° = −cos 105° · 3. sin⁻¹(0.8) = 53.13…° so 53° and 180° − 53.13° = 126.87…°, which is 127°

    Next: trigonometric equations →

    Solving trigonometric equations Extended

    🔑 Key ideas

    • Press the inverse button to get the acute reference angle α, using the value without its minus sign.
    • Use the table to find the solutions in 0° ≤ θ ≤ 360°.
    • If the question says an angle is obtuse (between 90° and 180°), pick the solution in that range.
    • A value outside −1 to 1 has no solutions for sin or cos.
    EquationValue positiveValue negative
    sin θ = kα and 180° − α180° + α and 360° − α
    cos θ = kα and 360° − α180° − α and 180° + α
    tan θ = kα and 180° + α180° − α and 360° − α
    180° − α sin positive α all positive 180° + α tan positive 360° − α cos positive 90°–180° 270°–360°
    The four quarters of 0°–360° for an acute reference angle α: in each quarter, the ratios named there are positive.

    📘 Example 22 · A is obtuse and sin A = . Find A

    1. The calculator gives sin−1() = 60°.
    2. But A is obtuse (between 90° and 180°), so 60° is not allowed.
    3. Use sin x = sin (180° − x): 180° − 60° = 120°.
    4. A = 120°. Check: sin 120° = sin 60° = . ✔

    📘 Example 23 · tan θ = for 0° ≤ θ ≤ 360°

    1. tan−1() = 30°.
    2. Tangent has period 180°, so add 180°: 30° + 180° = 210°.
    3. θ = 30° or 210°.

    📘 Example 24 · cos θ = −0.5 for 0° ≤ θ ≤ 360°

    1. Ignore the minus: α = cos−1(0.5) = 60°.
    2. Cosine is negative, so the table gives 180° − α and 180° + α.
    3. 180° − 60° = 120° and 180° + 60° = 240°.
    4. θ = 120° or 240°. Check: cos 240° = −cos 60° = −0.5. ✔

    📘 Example 25 · 5 cos θ + 1 = 2 for 0° ≤ θ ≤ 360°

    1. Subtract 1: 5 cos θ = 1.
    2. Divide by 5: cos θ = 0.2.
    3. α = cos−1(0.2) = 78.46…°. Cosine is positive: α and 360° − α.
    4. 360° − 78.46° = 281.54°.
    5. θ = 78.5° or 281.5° (1 d.p.).

    📘 Example 26 · tan θ = . Find sin θ and cos θ without a calculator

    1. Draw a right-angled triangle with opposite 3 and adjacent 4.
    2. Hypotenuse: √(3² + 4²) = √25 = 5.
    3. sin θ = O ÷ H = .
    4. cos θ = A ÷ H = .

    📘 Example 27 · Why does sin θ = cos θ when θ = 45°?

    1. sin θ = O ÷ H and cos θ = A ÷ H, with the same H.
    2. They are equal exactly when O = A.
    3. A right-angled triangle with O = A is isosceles, so the two acute angles are equal.
    4. They add to 90°, so each is 45°.

    ✏️ Try it

    1. Solve sin θ = 0.5 for 0° ≤ θ ≤ 360°.
    2. Solve tan θ = −1 for 0° ≤ θ ≤ 360°.
    3. cos θ = . Find sin θ without a calculator (θ acute).
    Answers

    1. α = 30°, sine positive: 30° and 150° · 2. α = 45°, tangent negative: 180° − 45° = 135° and 360° − 45° = 315° · 3. Adjacent 4, hypotenuse 5, so opposite 3: sin θ =

    Next: summary and quiz →

    Chapter summary

    • Label the sides from the chosen angle: Opposite, Adjacent, Hypotenuse.
    • SOH CAH TOA: sin θ = O ÷ H, cos θ = A ÷ H, tan θ = O ÷ A.
    • Finding a side: rearrange the ratio. Finding an angle: use sin−1, cos−1 or tan−1.
    • Pythagoras: a² + b² = c². Use it for two sides and no angle.
    • Angles of elevation and depression are measured from the horizontal. They are equal as alternate angles.
    • Extended sin x = sin (180° − x), cos x = −cos (180° − x). Sine and cosine repeat every 360°, tangent every 180°.
    • Extended For an equation, find the reference angle, then use the quarter table for solutions in 0° to 360°.

    📄 Practice questions

    The book's Exercises 29.1 to 29.9 and Student assessments 1, 2 and 3 are in the printable assignment for this chapter. Because the book's diagrams can't be printed, each is described in words.

    Mini-quiz

    1. Which ratio is sin θ in a right-angled triangle?
    2. A right-angled triangle has opposite 5 cm and adjacent 12 cm. What is the angle θ?
    3. The hypotenuse is 10 cm and an angle is 30°. How long is the side opposite that angle?
    4. The side next to a 60° angle is 8 cm. How long is the hypotenuse?
    5. A right-angled triangle has hypotenuse 17 cm and one shorter side 8 cm. The other side is…
    6. You are 40 m from a tree and its top is at an angle of elevation of 25°. How tall is the tree?
    7. sin 40° = 0.643. Which other angle between 0° and 180° also has a sine of 0.643?
    8. Which of these is equal to cos 70°?
    9. Solve tan θ = 1 for 0° ≤ θ ≤ 360°.
    10. Which equation has no solutions at all?
    11. From a cliff top, a boat is seen at an angle of depression of 20°. The cliff top from the boat is at…

    Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 29 (Trigonometry). Explanations, the Try it answers, the triangle solver and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.