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.
55–60 minMS-M2 — MEDIUM3 MC3 SALesson 14 of 22Free
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Choose how you work: type answers on screen, or work in your book.
Think First
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?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Write your initial thinking in your book
Saved
Come back to this at the end of the lesson.
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Trigonometric Ratios — This Lesson
$\sin\theta = \dfrac{\text{O}}{\text{H}}$
Sine — Opposite over HypotenuseUse when you have O and H (or need one of them), with a known angle
$\cos\theta = \dfrac{\text{A}}{\text{H}}$
Cosine — Adjacent over HypotenuseUse when you have A and H (or need one of them), with a known angle
$\tan\theta = \dfrac{\text{O}}{\text{A}}$
Tangent — Opposite over AdjacentUse when you have O and A (or need one of them), with a known angle; hypotenuse not involved
SOH — CAH — TOA
Memory aid — label H (hypotenuse) first, then O (opposite to angle) and A (adjacent) relative to the marked angle
🧠 Know
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
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
Label O, A, H for any angle in a right-angled triangle
Choose the correct trig ratio and solve for an unknown side
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.
Misconceptions to Fix
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Wrong: You can use any trig ratio to find an unknown side — just choose whichever you prefer
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Right: The correct ratio depends on which sides are known and unknown relative to the given angle. Label the sides as opposite, adjacent, and hypotenuse first, then choose the ratio that connects the known and unknown side.
Step 1 — Label and Choose
Key Point
Always check your units before substituting into formulas. Converting to consistent units is a common source of errors in assessment tasks.
Key Terms
FormulaA rule showing the relationship between variables using symbols.
SubstitutionReplacing variables with their known values in an equation.
Unit ConversionChanging a measurement from one unit to another.
CapacityThe amount of liquid a container can hold, measured in litres or millilitres.
PerimeterThe total distance around the outside of a shape.
AreaThe amount of space inside a two-dimensional shape.
Always Label First
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 $\sin\theta = \frac{5}{x}$, multiply both sides by $x$: $x \sin\theta = 5$, then $x = \frac{5}{\sin\theta}$.
Worked Example 1Unknown in Numerator
Problem
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.
Solution
1Known: H $= 18$ m, $\theta = 38°$. Need: O. Use $\sin\theta = \dfrac{O}{H}$O and H are involved; no A needed → use sine
2$O = H \times \sin\theta = 18 \times \sin 38°$Multiply both sides by H; unknown is in the numerator position
A right-angled triangle has angle 52° and the adjacent side is 9.4 cm. Find the hypotenuse, correct to 2 decimal places.
Solution
1Known: A $= 9.4$ cm, $\theta = 52°$. Need: H. Use $\cos\theta = \dfrac{A}{H}$A and H are involved → use cosine
2$\cos 52° = \dfrac{9.4}{H}$; multiply: $H \cos 52° = 9.4$; so $H = \dfrac{9.4}{\cos 52°}$Unknown is in denominator — multiply across then divide; OR: $H = A \div \cos\theta$
From a point 24 m from the base of a vertical tree, the angle of elevation to the top is 35°. Find the height of the tree, correct to 1 decimal place.
Solution
1Draw the triangle: horizontal base $= 24$ m (A), vertical height $= h$ (O), angle $= 35°$The horizontal distance is adjacent; the height is opposite the 35° angle
2$\tan 35° = \dfrac{h}{24}$; so $h = 24 \times \tan 35°$O and A are involved (no H needed) → use tangent; unknown in numerator
3$h = 24 \times 0.7002... \approx 16.8 \text{ m}$$\tan 35° \approx 0.7002$; round to 1 decimal place
Worked Example 4Degrees and Minutes
Problem
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.
Solution
1$63°24' = 63 + \dfrac{24}{60} = 63.4°$Convert degrees and minutes to decimal degrees: divide minutes by 60
2Known: O $= 4.6$ m (wall height), $\theta = 63.4°$. Need: H (ladder). Use $\sin\theta = \dfrac{O}{H}$Angle at base, wall is opposite, ladder is hypotenuse → sine
3$H = \dfrac{O}{\sin\theta} = \dfrac{4.6}{\sin 63.4°} \approx \dfrac{4.6}{0.8942} \approx 5.15 \text{ m}$Unknown in denominator: $H = O \div \sin\theta$
Practice Questions
For each question: label O, A, H; state the ratio chosen; show substitution and evaluation.
Section A — Standard Right-Angled Triangles
$\theta = 42°$, H $= 15$ cm. Find O (to 2 d.p.).
$\theta = 28°$, H $= 22$ m. Find A (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.).
$\theta = 71°$, A $= 5.5$ m. Find H (to 2 d.p.).
$\theta = 48°$, O $= 12$ cm. Find A (to 2 d.p.).
Section B — Degrees and Minutes
$\theta = 35°30'$, H $= 20$ m. Find O (to 2 d.p.).
$\theta = 47°12'$, A $= 6.3$ m. Find H (to 2 d.p.).
Section C — Practical Problems
A ramp makes an angle of 15° with the horizontal. If the ramp is 8 m long, find its vertical rise.
A flagpole casts a shadow of 12 m. The sun's angle of elevation is 54°. Find the height of the flagpole (to 1 d.p.).
A cable car wire goes from ground level to a station 420 m higher. The wire makes an angle of 28° with the horizontal. Find the length of the wire (to the nearest metre).
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.).
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
1 In a right-angled triangle, the angle is 34° and the hypotenuse is 20 cm. The side adjacent to the 34° angle is closest to:
A 11.2 cm
B 13.6 cm
C 16.6 cm
D 35.9 cm
C — $A = 20\cos34° \approx 20 \times 0.829 = 16.6$ cm.
2 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:
A 4.11 m
B 9.64 m
C 5.35 m
D 4.97 m
B — $h = 4.5\tan65° \approx 4.5 \times 2.145 = 9.64$ m.
3 From a point on the ground, the angle of elevation to the top of a 60 m building is 41°. The distance from the point to the base of the building is closest to:
A 39.4 m
B 52.3 m
C 69.0 m
D 79.6 m
C — $d = 60/\tan41° \approx 60/0.8693 \approx 69.0$ m.
Short Answer
01
SA 43 marks
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.). (2 marks)
(b) Find the distance from the base of the wall to the foot of the ladder (to 2 d.p.). (1 mark)
SA 64 marks
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. (1 mark)
Find how far east the ship has sailed, correct to 2 decimal places. (2 marks)
Find the direct distance from the ship to the lighthouse, correct to 2 decimal places. (1 mark)
Work in your book
Saved
Diagram
Right angle at the ship's position; north side = 3.2 km (O); east distance = A; angle at ship = 25° (bearing N25°E means 25° from north towards east); hypotenuse = direct distance to lighthouse
East distance
$\tan25° = \frac{A}{3.2}$... wait: the angle at the ship between north and the lighthouse direction is 25°; so O is the east distance, A is the north distance (3.2 km). $A_{\text{east}} = 3.2\tan25° \approx 3.2 \times 0.4663 \approx \mathbf{1.49 \text{ km}}$