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Module 2 · L4 of 22 ~55 min ⚡ +95 XP available

Introduction to Trigonometry

Label the triangle, select the ratio, solve the equation. Three steps. Every trigonometry problem in this course follows this pattern — SOHCAHTOA unlocks them all.

Today's hook — Two right-angled triangles both have a 35° angle, but one is much larger. The ratio of opposite to hypotenuse is the same in both. Why? And how does that let us find unknown sides?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall — your gut answer first
+5 XP warm-up

Two right-angled triangles both have a 35° angle, but one is much larger than the other. Would the ratio of the opposite side to the hypotenuse be the same in both triangles, or different? Why?

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Come back to this at the end of the lesson.

02
The three ratios you need to own — SOHCAHTOA
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For any right-angled triangle with a given angle $\theta$, the ratios of pairs of sides are fixed — they depend only on the angle, not the size of the triangle. SOHCAHTOA names these three ratios.

SOH: $\sin\theta = \dfrac{O}{H}$ — Opposite over Hypotenuse

CAH: $\cos\theta = \dfrac{A}{H}$ — Adjacent over Hypotenuse

TOA: $\tan\theta = \dfrac{O}{A}$ — Opposite over Adjacent

θ O A H
O and H identified → use sin  ·  A and H → use cos  ·  O and A → use tan
SOH — sinθ = O/H
Rearranged: $O = H\sin\theta$ (multiply) or $H = O/\sin\theta$ (divide). Use when opposite and hypotenuse are involved.
CAH — cosθ = A/H
Rearranged: $A = H\cos\theta$ or $H = A/\cos\theta$. Use when adjacent and hypotenuse are involved.
TOA — tanθ = O/A
Rearranged: $O = A\tan\theta$ or $A = O/\tan\theta$. Use when opposite and adjacent are involved (no hypotenuse).
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What you'll master
Know

Key facts

  • The three trig ratios — sine, cosine, tangent — and their abbreviations
  • The SOHCAHTOA memory device
  • How to use $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ on a calculator
Understand

Concepts

  • Why the ratio of two sides depends only on the angle, not the triangle's size
  • Why selecting the ratio requires identifying the two sides involved
  • Why inverse trig functions find angles rather than sides
Can do

Skills

  • Label opposite, adjacent, and hypotenuse relative to any given angle
  • Select and apply the correct ratio to find an unknown side or angle
  • Use a calculator in degree mode correctly for all trig calculations
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Key terms
Opposite sideThe side directly across from the reference angle — not touching it.
Adjacent sideThe side next to the reference angle that is not the hypotenuse.
Sine (sin)The ratio of the opposite side to the hypotenuse for a given angle.
Cosine (cos)The ratio of the adjacent side to the hypotenuse for a given angle.
Tangent (tan)The ratio of the opposite side to the adjacent side for a given angle.
Degree modeCalculator setting required for all HSC trig — verify by checking sin(90) = 1.
05
Why trigonometry works — and how to label any triangle
core concept

Take any right-angled triangle with a 35° angle. Make it bigger. Make it smaller. As long as that angle stays at 35°, the ratio of any two sides stays exactly the same. That number is $\sin 35°$, $\cos 35°$, or $\tan 35°$.

Trigonometry names and uses these fixed ratios. Once you know an angle, you know all three ratios. Once you know a ratio and a side length, you can find any other side.

Why this is useful. Real-world triangles — ramps, roofs, survey lines, staircases — can be described by an angle and one side. Trig finds the other sides without measuring them physically.

Labelling the triangle — the most important step

Before writing any formula, label the three sides relative to the angle you are working with. This is the step most often skipped — and the one that causes the most errors.

Labelling process (5 seconds, every time):

  1. Mark the right angle
  2. Circle the reference angle (the one you know or are finding)
  3. Label H opposite the right angle
  4. Label O opposite the circled angle
  5. Label A — the remaining side
LABELLING CHANGES WITH THE REFERENCE ANGLE θ O A H θ at bottom-right θ H O A θ at top-left — O and A swap! H never changes. O and A always do. LABELLING THE TRIANGLE — SOHCAHTOA θ O opp A adj H hyp SOH sinθ = O/H opp ÷ hyp CAH cosθ = A/H adj ÷ hyp TOA tanθ = O/A opp ÷ adj Which ratio to use? O & H → sin  ·  A & H → cos  ·  O & A → tan Identify the two sides involved, then match the ratio

Finding an unknown angle — inverse trig

When you know two sides and need the angle, use the inverse trig functions:

$$\text{If } \sin\theta = 0.6, \text{ then } \theta = \sin^{-1}(0.6)$$

On your calculator: press SHIFT (or 2ND) then the sin/cos/tan key.

Degrees and minutes: HSC questions sometimes require angles in degrees and minutes. Multiply the decimal part of the angle by 60 to convert to minutes. For example: $\theta = 34.7°$ → $0.7 \times 60 = 42$ min → $\theta = 34°42'$.

Algebra beats memory. Never try to remember "multiply or divide" as a rule. Always write the equation ($\sin\theta = O/H$) and solve algebraically. Two lines of working: rearrange, then evaluate. This earns method marks even if the final calculation goes wrong.
What to write in your book
  • SOHCAHTOA: sin = O/H · cos = A/H · tan = O/A
  • Step 1 — always label O, A, H relative to the reference angle before writing any formula.
  • Step 2 — identify the two sides involved; match to the ratio table.
  • Step 3 — write the equation, rearrange algebraically, evaluate.
  • Finding an angle: use $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$ (SHIFT + trig key). Confirm degree mode: $\sin(90) = 1$.
  • Degrees and minutes: multiply decimal degrees by 60 for minutes. $34.7° = 34°42'$.

Quick check: The reference angle is 40°. The known side is the hypotenuse. The unknown side is the opposite. Which ratio should you use?

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FINDING AN UNKNOWN SIDE (OPPOSITE)

In a right-angled triangle, the reference angle is 32°, the hypotenuse is 15 cm, and $x$ is the opposite side. Find $x$ correct to 2 decimal places.

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Label: O = $x$ (unknown), H = 15 cm (known) → sides O and H → use sin
Label the triangle first: circle 32°, label H opposite the right angle, label O opposite 32°. Two sides involved: opposite and hypotenuse.
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PROBLEM 2 · FINDING THE HYPOTENUSE

In a right-angled triangle, the reference angle is 48°, the adjacent side is 9 m, and the hypotenuse is unknown. Find the hypotenuse correct to 2 decimal places.

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Label: A = 9 m (known), H = unknown → sides A and H → use cos
Adjacent and hypotenuse — use cosine.
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PROBLEM 3 · FINDING AN ANGLE IN DEGREES AND MINUTES

Find the angle $\theta$ if $\tan\theta = 0.842$. Give your answer in degrees and minutes.

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$\theta = \tan^{-1}(0.842) = 40.0876...°$
Given the tan ratio directly — apply $\tan^{-1}$ immediately. Calculator in degree mode.
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What to write in your book
  • When unknown is in the numerator: multiply. When in denominator: rearrange algebraically (two lines).
  • Finding an angle: $\theta = \sin^{-1}(O/H)$ — leave the fraction inside the inverse function for better precision.
  • Degrees and minutes: decimal part × 60 = minutes. E.g. $28.0724° → 0.0724 \times 60 = 4.3 \approx 4'$ → $28°4'$.

Did you get this? True or false: when the unknown side is in the denominator of a trig equation, you solve by multiplying both sides by the unknown, then dividing by the trig ratio.

Common errors · the 3 traps that cost marks

Trap 01
Wrong ratio because triangle wasn't labelled first
Guessing the ratio based on the shape of the triangle rather than identifying the sides. Fix: label O, A, H relative to the reference angle on every diagram before writing any formula. 5 seconds. Non-negotiable.
Trap 02
Calculator in radian mode
$\sin(32)$ in radian mode gives $-0.5514$ instead of $0.5299$. The answer will be completely wrong. Check: $\sin(90) = 1$ before every trig calculation.
Trap 03
Wrong rearrangement when unknown is in the denominator
$\cos 48° = \dfrac{9}{H}$ becomes $H = 9 \times \cos 48°$ (wrong — multiplied instead of divided). Always solve algebraically: multiply both sides by $H$, then divide both sides by $\cos 48°$. Write both lines explicitly.
What to write in your book
  • Always label first — O, A, H relative to the reference angle. This determines the ratio. Takes 5 seconds.
  • Degree mode check: $\sin(90) = 1$. Do this before every trig session and exam.
  • Unknown in denominator: two algebra steps — multiply by unknown, divide by trig ratio. Never skip steps.

Fill the gap: A reference angle of 55° with adjacent = 10 m and opposite = unknown. The ratio to use is , giving $O = 10 \times \tan 55° \approx$ m.

Quick-fire practice · 13 calculations

1

Reference angle = 40°, known = hypotenuse, unknown = opposite. Which ratio?

2

Reference angle = 55°, known = adjacent, unknown = hypotenuse. Which ratio?

3

Reference angle = 28°, known = opposite, unknown = adjacent. Which ratio?

4

Reference angle = 63°, known = hypotenuse, unknown = adjacent. Which ratio?

5

Reference angle = 35°, hypotenuse = 12 cm. Find the opposite side.

6

Reference angle = 50°, hypotenuse = 20 m. Find the adjacent side.

7

Reference angle = 42°, adjacent = 8 cm. Find the hypotenuse.

8

Reference angle = 67°, opposite = 14 m. Find the hypotenuse.

9

Reference angle = 38°, adjacent = 10 cm. Find the opposite side.

10

Reference angle = 22°, opposite = 5 m. Find the adjacent side.

11

Opposite = 6 cm, hypotenuse = 10 cm. Find $\theta$ to the nearest degree.

12

Adjacent = 8 m, hypotenuse = 13 m. Find $\theta$ to the nearest degree.

13

Opposite = 9 cm, adjacent = 4 cm. Find $\theta$ in degrees and minutes.

Odd one out: Three of these are correct statements about trigonometry. Which one is wrong?

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Revisit your thinking

Earlier you predicted whether the ratio of opposite to hypotenuse would be the same in two right-angled triangles both having a 35° angle. The answer is yes — always the same.

For any 35° right-angled triangle in the world, $O \div H = \sin 35° \approx 0.5736$. This is why trigonometry works: the ratio is a property of the angle alone, not the triangle's size. That's the entire basis of the subject.

Look back at what you wrote. What has changed? What did you get right? What surprised you?

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 42 marks

Q1. Find the length of side $x$ in a right-angled triangle where the reference angle is 41°, $x$ is the adjacent side, and the hypotenuse is 18 cm. Give your answer correct to 2 decimal places. (2 marks)

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ApplyBand 43 marks

Q2. A right-angled triangle has opposite side 8 m and hypotenuse 17 m. (a) Write down the trigonometric ratio that connects these two sides. (1 mark) (b) Find the reference angle $\theta$ correct to the nearest minute. (2 marks) (3 marks total)

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AnalyseBand 54 marks

Q3. A ramp rises from ground level to a platform. The ramp makes an angle of 15° with the horizontal. The horizontal distance is 6 m. (a) Draw a labelled diagram showing this situation. (1 mark) (b) Find the length of the ramp (hypotenuse) correct to 2 decimal places. (1 mark) (c) Find the height of the platform correct to 2 decimal places. (2 marks) (4 marks total)

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📖 Comprehensive answers (click to reveal)

Drill 1: sin  ·  2: cos  ·  3: tan  ·  4: cos  ·  5: $O = 12\sin 35° = 6.88\text{ cm}$  ·  6: $A = 20\cos 50° = 12.86\text{ m}$  ·  7: $H = 8/\cos 42° = 10.76\text{ cm}$  ·  8: $H = 14/\sin 67° = 15.21\text{ m}$  ·  9: $O = 10\tan 38° = 7.81\text{ cm}$  ·  10: $A = 5/\tan 22° = 12.37\text{ m}$  ·  11: $37°$  ·  12: $52°$  ·  13: $66°2'$

Q1 (2 marks): $\cos 41° = x/18$; $x = 18\cos 41° = \mathbf{13.58\text{ cm}}$ [2].

Q2 (3 marks): (a) $\sin\theta = O/H = 8/17$ [1]. (b) $\theta = \sin^{-1}(8/17) = 28.0724...°$; $0.0724 \times 60 = 4.3 \approx 4'$; $\theta = \mathbf{28°4'}$ [2].

Q3 (4 marks): (a) Sketch showing right angle at top of vertical, 15° at base, horizontal = 6 m [1]. (b) $\cos 15° = 6/H$; $H = 6/\cos 15° = \mathbf{6.21\text{ m}}$ [1]. (c) $\tan 15° = \text{height}/6$; height $= 6\tan 15° = \mathbf{1.61\text{ m}}$ [2].

01
Boss battle · The Angle Master
earn bronze · silver · gold

Five timed questions on trigonometry. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering trigonometry questions. Pool: lesson 4.

Mark lesson as complete

Tick when you've finished the practice and review.

Module overview · Year 11 Maths Standard