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.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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?
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.
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
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
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
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
What to write in your book
- SOH: $\sin\theta = \frac{O}{H}$ — opposite over hypotenuse. Use when O and H involved.
- CAH: $\cos\theta = \frac{A}{H}$ — adjacent over hypotenuse. Use when A and H involved.
- TOA: $\tan\theta = \frac{O}{A}$ — opposite over adjacent. Use when O and A involved (no H needed).
- Unknown in numerator: side = H (or A) × trig ratio. Unknown in denominator: side = known side ÷ trig ratio.
Did you get this? True or false: in a right-angled triangle, the hypotenuse is the side opposite the marked angle $\theta$.
Worked examples · 3 in a row, reveal as you go
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.
A right-angled triangle has angle 52° and the adjacent side is 9.4 cm. Find the hypotenuse, correct to 2 decimal places.
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.
What to write in your book
- Degrees and minutes: $a°b' = a + \frac{b}{60}$ degrees (decimal). Always convert before calculator entry.
- Angle of elevation (looking up) = angle inside triangle above horizontal.
- Angle of depression (looking down) = angle inside triangle below horizontal = alternate angle inside triangle.
- Practical shortcut: if unknown is in numerator, multiply; if in denominator, divide.
Quick check: A right-angled triangle has $\theta = 34°$ and hypotenuse $= 20$ cm. The side adjacent to the 34° angle is closest to:
Common errors · the 3 traps that cost marks
What to write in your book
- Always re-label O, A, H for the specific angle in each question.
- Degrees and minutes: convert before calculator. $a°b' = (a + b/60)°$.
- Angle of depression = angle of elevation (alternate angles).
- Check: H must be the longest side. If your answer for H is shorter than O or A, you've made an error.
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($$°)$.
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:
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
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)
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)
📖 Comprehensive answers (click to reveal)
Drill 1: $O = 15\sin42° \approx 10.04$ cm · 2: $O = 8.4\tan55° \approx 12.00$ cm · 3: $H = 7/\sin36° \approx 11.90$ m · 4: $35°30' = 35.5°$; $O = 20\sin35.5° \approx 11.60$ 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].
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.
⚔ Enter the arenaClimb platforms by answering right-angled trigonometry questions. Pool: lesson 14.
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