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Module 2 · L14 of 22 55–60 min ⚡ +95 XP available

Right-Angled Trigonometry: Finding Unknown Sides

Label the triangle, choose the correct ratio, then solve, three consistent steps every time. When the unknown is in the denominator, multiply across. When it is in the numerator, the answer falls straight out.

Today's hook, A ramp rises 1.2 m vertically over a horizontal distance. You need to find the length of the ramp surface. What information would you need? What shape does this form? Which trigonometric ratio would connect the angle of the ramp to the length you need?

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Orient to unknown sides

Recall the three ratios, set the goal and settle the key terms.

Worksheets

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

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Recall, your gut answer first

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A ramp rises 1.2 m vertically over a horizontal distance. You need to find the length of the ramp surface.

Without calculating what information would you need? What shape does this form? Which trigonometric ratio would connect the angle of the ramp to the length you need?

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The three ratios you need to own

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Trigonometry uses three core ratios. Every side-finding question is just a selection from these three and then a rearrangement.

Label first H (hypotenuse, opposite right angle), O (opposite the marked angle), A (adjacent to the marked angle). Then choose the ratio that involves the two sides you have or need. Only then substitute and solve.

Label the sides relative to the angle, then pick the ratio that uses the two you have.

SOH · CAH · TOA, label H first, then O and A relative to $\theta$

Unknown in numerator

$\sin\theta = \frac{O}{H}$ → $O = H \times \sin\theta$. Multiply both sides by H. Answer falls straight out.

Unknown in denominator

$\cos\theta = \frac{A}{H}$ → $H = \frac{A}{\cos\theta}$. Multiply across then divide. A common source of errors.

Degrees and minutes

$63°24' = 63 + \frac{24}{60} = 63.4°$. Always convert to decimal degrees before using your calculator.

What you'll master

Know

Key facts

SOHCAHTOA, the three ratios and when to use each
H is always opposite the right angle; O and A depend on which angle is marked
When unknown is in the numerator: multiply; when in denominator: multiply across
Angles given in degrees and minutes, convert to decimal degrees for calculator

Understand

Concepts

Why labelling the triangle before choosing a ratio prevents errors
How to handle the two cases: unknown in numerator vs denominator
How trig applies in real-world elevation/depression problems

Can do

Skills

Label O, A, H for any angle in a right-angled triangle
Choose the correct trig ratio and solve for an unknown side
Solve multi-step practical problems (ladders, ramps, bearings)
Round answers appropriately and include units

Key terms

Hypotenuse (H)The longest side; always opposite the right angle; used in both sin and cos ratios.

Opposite (O)The side directly across from the marked angle (not touching it); used in sin and tan.

Adjacent (A)The side next to the marked angle that is not the hypotenuse; used in cos and tan.

Angle of elevation/depressionElevation: angle measured upward from horizontal; depression: angle measured downward from horizontal. Both create right-angled triangles.

Label before you solve

Fix the reference angle first, then name the two sides the question involves.

Always label first

core concept

Before writing any formula, label the three sides of your right-angled triangle relative to the angle given. This one step prevents almost all SOHCAHTOA errors.

Mark the right angle with a small square
Label H (hypotenuse), always opposite the right angle
Mark the given angle $\theta$
Label O, side opposite $\theta$ (does not touch $\theta$)
Label A, side adjacent to $\theta$ (touches $\theta$, not the hypotenuse)
Choose the ratio that involves the two sides you have or need

Unknown in denominator, If $\sin\theta = \dfrac{5}{x}$, multiply both sides by $x$: $x \sin\theta = 5$, then $x = \dfrac{5}{\sin\theta}$.

SOH-CAH-TOA: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent. Always label the three sides first (relative to the given angle) before selecting and applying a ratio.

Pause, copy all three SOH-CAH-TOA ratios in full: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse, tan θ = opposite/adjacent, and write the three-step procedure: label sides, identify the known and unknown sides, then choose the matching ratio into your book.

Did you get this? True or false: in a right-angled triangle, the hypotenuse is the side opposite the marked angle $\theta$.

Work three side examples

Follow an unknown in the numerator, an unknown in the denominator and an applied elevation problem.

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

PROBLEM 1 · UNKNOWN IN NUMERATOR

In a right-angled triangle, the hypotenuse is 18 m and one angle is 38°. Find the side opposite the 38° angle, correct to 2 decimal places.

Known: H $= 18$ m, $\theta = 38°$. Need: O. Use $\sin\theta = \dfrac{O}{H}$

O and H are involved; no A needed → use sine (SOH).

PROBLEM 2 · UNKNOWN IN DENOMINATOR

A right-angled triangle has angle 52° and the adjacent side is 9.4 cm. Find the hypotenuse, correct to 2 decimal places.

Known: A $= 9.4$ cm, $\theta = 52°$. Need: H. Use $\cos\theta = \dfrac{A}{H}$

A and H are involved → use cosine (CAH).

PROBLEM 3 · ANGLE OF ELEVATION & DEGREES/MINUTES

A ladder makes an angle of 63°24' with the ground. If the ladder reaches 4.6 m up a vertical wall, find the length of the ladder, correct to 2 decimal places.

$63°24' = 63 + \dfrac{24}{60} = 63.4°$

Convert degrees and minutes to decimal degrees: divide minutes by 60.

Quick check: A right-angled triangle has $\theta = 34°$ and hypotenuse $= 20$ cm. The side adjacent to the 34° angle is closest to:

Avoid the three traps

Spot mislabelling, multiplying when you should divide, and minutes conversion slips.

Common errors · the 3 traps that cost marks

Trap 01

Labelling O and A relative to the wrong angle

O is opposite the marked angle, not the right angle. If a question uses the other acute angle instead, your O and A swap. Always re-label for the angle given in the question, this is the single biggest source of SOHCAHTOA errors.

Trap 02

Forgetting to convert degrees and minutes

$\sin(63°24')$ typed directly gives the wrong answer. Convert first: $63°24' = 63.4°$. Check the question for the ° ′ notation, it's common in HSC exams.

Trap 03

Treating angle of depression incorrectly

The angle of depression from the top equals the angle of elevation from the bottom (alternate angles). Both give the same angle inside the right-angled triangle do not subtract from 90°.

Fill the gap: $47°12'$ converted to decimal degrees is $47 + \dfrac{12}{60} = 47.$°. A triangle with $\theta = 47.2°$ and $A = 6.3$ m uses cosine: $H = 6.3 \div \cos($$°)$.

Drill unknown sides

Run the quick-fire calculations until the setup is automatic.

Quick-fire practice · 5 calculations

$\theta = 42°$, H $= 15$ cm. Find O (to 2 d.p.).

$\theta = 55°$, A $= 8.4$ cm. Find O (to 2 d.p.).

$\theta = 36°$, O $= 7$ m. Find H (to 2 d.p.).

$\theta = 35°30'$, H $= 20$ m. Find O (to 2 d.p.).

From the top of a 25 m cliff, the angle of depression to a boat is 18°. Find the horizontal distance from the base of the cliff to the boat (to 1 d.p.).

Quick match: A vertical pole has a wire attached to its top. The wire makes an angle of 65° with the ground and is anchored 4.5 m from the base. The height of the pole is closest to:

Revisit your thinking

Return to your opening answer and name what has changed.

Revisit your thinking

Earlier you considered finding the length of a ramp that rises 1.2 m vertically. Let's work it through:

The shape is a right-angled triangle. The vertical rise (1.2 m) is the side opposite the angle of inclination. The ramp surface is the hypotenuse. So you would use SOH: $\sin\theta = \frac{O}{H} = \frac{1.2}{H}$, giving $H = \frac{1.2}{\sin\theta}$.

For example, if the angle is 15°: $H = \frac{1.2}{\sin 15°} \approx \frac{1.2}{0.2588} \approx 4.63$ m.

You would also need the angle of inclination one piece of information beyond the 1.2 m rise.

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Practise finding sides

Answer the question bank, then write full short-answer responses.

Multiple choice

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

Short answer

ApplyBand 43 marks

Q1. A 6-metre ladder leans against a vertical wall. The ladder makes an angle of 72° with the ground. (a) Find how high up the wall the ladder reaches (to 2 d.p.). (b) Find the distance from the base of the wall to the foot of the ladder (to 2 d.p.). (3 marks)

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ApplyBand 43 marks

Q2. A surveyor stands 80 m from the base of a vertical cliff. She measures the angle of elevation to the top of the cliff as 38°42'. (a) Convert 38°42' to decimal degrees. (b) Find the height of the cliff, correct to the nearest metre. (3 marks)

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AnalyseBand 54 marks

Q3. A ship sails due east from port. After some time, the navigator observes a lighthouse at a bearing of N25°E, directly ahead of the ship's original direction. The lighthouse is known to be 3.2 km north of port. Draw and label the right-angled triangle formed by port, the ship, and the lighthouse. Find how far east the ship has sailed (to 2 d.p.) and the direct distance from the ship to the lighthouse (to 2 d.p.). (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $O = 15\sin42° \approx 10.04$ cm · 2: $O = 8.4\tan55° \approx 12.00$ cm · 3: $H = 7/\sin36° \approx 11.91$ m · 4: $35°30' = 35.5°$; $O = 20\sin35.5° \approx 11.61$ m · 5: $d = 25/\tan18° \approx 76.9$ m

Q1 (3 marks): (a) $h = 6\sin72° \approx 6 \times 0.9511 \approx 5.71$ m [2]. (b) $d = 6\cos72° \approx 6 \times 0.3090 \approx 1.85$ m [1].

Q2 (3 marks): (a) $38 + 42/60 = 38.7°$ [1]. (b) $h = 80\tan38.7° \approx 80 \times 0.8002 \approx 64$ m [2].

Q3 (4 marks): Diagram: right angle at the ship's position; north side = 3.2 km; angle at ship = 25° (bearing N25°E means 25° from north towards east) [1]. East distance: $A_{\text{east}} = 3.2\tan25° \approx 3.2 \times 0.4663 \approx 1.49$ km [2]. Direct distance: $H = 3.2/\cos25° \approx 3.2/0.9063 \approx 3.53$ km [1].

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Five timed questions on right-angled trigonometry: finding unknown sides. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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← Lesson 1 · Introduction to Trigonometry Lesson 3 · Finding Unknown Angles →

Focus area overview · Year 12 · Maths Standard