Pythagoras' theorem and trigonometry
Finding a missing side with Pythagoras' theorem, and the sine, cosine and tangent ratios
About four lessons of 45 to 60 minutes
How tall is that tree, without climbing it?
You cannot climb a tall tree with a tape measure, but you can measure how far you are standing from its base, and the angle up to its top. That is enough. Right-angled triangles hide inside almost every measuring problem where climbing, reaching or getting close is not an option: tree heights, building heights, the length of a zip line, how far out to sea a lighthouse can be seen.
Two tools unlock this. Pythagoras' theorem connects the three sides of a right-angled triangle, so knowing any two gives you the third. Trigonometric ratios, sine, cosine and tangent, connect an ANGLE to a ratio of sides, so knowing one side and one angle gives you the rest. Together they solve almost any right-triangle measuring problem.
- The height of a tree from its shadow's angletrigonometry finds height without climbing
- How far a ladder reaches up a wallPythagoras' theorem connects the ladder, wall and ground distance
- A lighthouse's line of sight to a shipthe angle of elevation and distance form a right triangle
- A ramp's length for a given rise and runthe same theorem used by builders every day
What students will be able to do
Students will use Pythagoras' theorem to find a missing side of a right-angled triangle, calculate the sine, cosine and tangent ratios for a given angle, explain why these ratios stay constant for similar triangles, and apply both tools to solve real-world spatial problems.
- I can use Pythagoras' theorem (a^2 + b^2 = c^2) to find the hypotenuse of a right-angled triangle.
- I can rearrange Pythagoras' theorem to find a missing leg when the hypotenuse is known.
- I can identify the opposite, adjacent and hypotenuse sides relative to a given angle.
- I can calculate sin, cos and tan as ratios of sides, and explain why they stay the same for any similar right-angled triangle.
- I can apply Pythagoras' theorem or a trigonometric ratio to solve a real-world measuring problem.
Standards this unit teaches
- AC9M9M03Australian Curriculum v9 (ACARA)Trigonometry and Pythagoras
Solve spatial problems by applying angle properties, scale, similarity, Pythagoras' theorem and right-angled trigonometry.
- AC9M9SP01Australian Curriculum v9 (ACARA)Trigonometric ratios
Recognise that the sine, cosine and tangent ratios stay constant for a given angle in right-angled triangles because of similarity.
Prior knowledge
This unit builds on skills students should already have met. Revisit any that are shaky first.
- Year 7 linear equations & expressions teaching unitrearranging a formula (like solving for a missing leg) is the same skill as solving an equation
- Year 9 quadratics teaching unitPythagoras' theorem itself involves squares, the same idea used throughout quadratics
- Geometry worksheetsangle and shape basics this unit builds on
Words to teach and display
- Hypotenuse
- the longest side of a right-angled triangle, always opposite the right angle
- Leg
- either of the two shorter sides of a right-angled triangle, forming the right angle
- Opposite side
- the side directly across from the angle being considered, not touching it
- Adjacent side
- the shorter side that touches the angle being considered, but is not the hypotenuse
- Sine, cosine, tangent
- the three trigonometric ratios: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent
- Similar triangles
- triangles with the same angles and proportional sides, but not necessarily the same size
Teach it: concrete, pictorial, abstract
The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.
1. Pythagoras' theorem: finding the hypotenuse
ConcreteIn any right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: a^2 + b^2 = c^2, where c is always the hypotenuse.
This works for every right-angled triangle, any size. A triangle with legs 3 and 4 always has a hypotenuse of exactly 5, because 3^2 + 4^2 = 9 + 16 = 25, and the square root of 25 is 5.
A right-angled triangle has legs of length 3 and 4. Find the hypotenuse.
- Apply Pythagoras' theorem: c^2 = a^2 + b^2 = 3^2 + 4^2.
- 3^2 + 4^2 = 9 + 16 = 25.
- c = sqrt(25) = 5.
Answer: The hypotenuse is 5.
- Which side is always 'c' in a^2 + b^2 = c^2?
- Why must you take a square root as the last step?
2. Finding a missing leg
PictorialIf the hypotenuse and one leg are known instead, rearrange the theorem: a leg squared equals the hypotenuse squared minus the other leg squared.
For a hypotenuse of 13 and a leg of 5: the missing leg squared is 13^2 - 5^2 = 169 - 25 = 144, so the missing leg is the square root of 144, which is 12. Notice 5, 12, 13 is a whole-number 'Pythagorean triple', just like 3, 4, 5.
A right-angled triangle has a hypotenuse of 13 and one leg of 5. Find the other leg.
- Rearrange: missing leg^2 = hypotenuse^2 - known leg^2 = 13^2 - 5^2.
- 13^2 - 5^2 = 169 - 25 = 144.
- missing leg = sqrt(144) = 12.
Answer: The missing leg is 12.
- Why does finding a missing leg SUBTRACT the squares instead of adding them?
- How can you check 5, 12, 13 really is a right-angled triangle's sides?
3. Trigonometric ratios: SOH-CAH-TOA
AbstractFor an angle in a right-angled triangle (not the right angle itself), three ratios of sides are given special names: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent. The memory phrase SOH-CAH-TOA captures all three.
Crucially, these ratios depend only on the ANGLE, not the triangle's size. A 3-4-5 triangle and a 6-8-10 triangle (exactly double) have the same angles, because one is a scaled-up copy of the other (they are similar triangles), so their trig ratios match exactly: 3/5 = 6/10 = 0.6.
A right-angled triangle has legs 3 and 4 and hypotenuse 5. The angle theta is opposite the side of length 3. Find sin(theta), cos(theta) and tan(theta).
- sin(theta) = opposite/hypotenuse = 3/5 = 0.6.
- cos(theta) = adjacent/hypotenuse = 4/5 = 0.8.
- tan(theta) = opposite/adjacent = 3/4 = 0.75.
Answer: sin(theta) = 0.6, cos(theta) = 0.8, tan(theta) = 0.75.
- If a triangle is scaled up to twice its size, what happens to its trig ratios, and why?
- Which ratio, sin, cos or tan, does NOT involve the hypotenuse at all?
4. Applying it: real-world measuring problems
AbstractBack to the hook: a ladder, a tree, a lighthouse, each is a right-angled triangle in disguise. Identify which sides or angles you already know, choose Pythagoras (three sides) or a trig ratio (an angle plus sides), and solve.
A 17 m ladder leans against a wall with its base 8 m from the wall. Using Pythagoras (8, 15, 17 is a Pythagorean triple): the height up the wall is sqrt(17^2 - 8^2) = sqrt(289 - 64) = sqrt(225) = 15 m. The same 8-15-17 triangle also gives tan(theta) = 15/8 = 1.875 for the angle the ladder makes with the ground, no protractor needed.
- In the ladder problem, which side is the hypotenuse: the ladder, the wall, or the ground distance?
- How would you decide whether to use Pythagoras' theorem or a trig ratio for a new problem?
Common misconceptions and how to address them
MisconceptionThe hypotenuse can be any side you choose in a^2 + b^2 = c^2.
Why it happens: Students apply the formula mechanically without first identifying which side is opposite the right angle.
How to address it: The hypotenuse (c) is always the LONGEST side, always opposite the right angle. Identify it first, every time, before assigning letters to the other two sides.
MisconceptionTo find a missing leg, you add the squares just like finding the hypotenuse.
Why it happens: Students over-generalise the addition step without noticing which side is unknown.
How to address it: Only find the hypotenuse by ADDING the other two squares. If the hypotenuse is already known and a leg is missing, SUBTRACT the known leg's square from the hypotenuse's square instead.
MisconceptionA bigger triangle has bigger trig ratios, since its sides are bigger.
Why it happens: Students confuse the absolute side lengths growing with the RATIO between them changing.
How to address it: Scaling a triangle up multiplies every side by the same amount, so the RATIO between any two sides stays exactly the same. Show 3/5 and 6/10 both equal 0.6 side by side.
Guided practice (with answers)
1. A right-angled triangle has legs 6 and 8. Find the hypotenuse.
Answer: 10, because 6^2 + 8^2 = 36 + 64 = 100, and sqrt(100) = 10.
2. A right-angled triangle has a hypotenuse of 25 and one leg of 7. Find the other leg.
Answer: 24, because 25^2 - 7^2 = 625 - 49 = 576, and sqrt(576) = 24.
3. A right-angled triangle has legs 9 and 12 and hypotenuse 15. The angle theta is opposite the side of length 9. Find sin(theta).
Answer: 0.6, because sin(theta) = opposite/hypotenuse = 9/15 = 0.6.
4. Using the same triangle (legs 9 and 12, hypotenuse 15, theta opposite the side of length 9), find tan(theta).
Answer: 0.75, because tan(theta) = opposite/adjacent = 9/12 = 0.75.
5. A 10 m ladder leans against a wall with its base 6 m from the wall. How high up the wall does it reach?
Answer: 8 m, because 10^2 - 6^2 = 100 - 36 = 64, and sqrt(64) = 8.
Independent practice worksheets
Practise Pythagoras' theorem and trigonometric ratios with computed, never-wrong answer keys, using real Pythagorean triples so every answer is exact.
Differentiation
- Start with only 'find the hypotenuse' problems using the 3-4-5 triple before introducing missing-leg problems.
- Provide a labelled diagram template (hypotenuse, opposite, adjacent relative to a marked angle) for every new triangle.
- Keep a reference card of common Pythagorean triples (3-4-5, 5-12-13, 8-15-17) so the arithmetic stays clean while the concept is new.
- Physically overlay a 3-4-5 triangle and a scaled 6-8-10 triangle (cut from paper) to make the similarity/ratio connection tangible.
- Introduce triangles where the sides are not whole numbers, requiring a calculator for the square root.
- Ask students to derive the exact trig ratios for a 45-45-90 triangle and a 30-60-90 triangle from first principles.
- Explore angle of elevation and angle of depression problems, including two-step problems needing both Pythagoras and a trig ratio.
- Investigate why sin(theta) and cos(90 - theta) are always equal, using the two acute angles of the same right triangle.
Assessment: exit ticket
A three-question exit ticket sampling Pythagoras' theorem (both directions) and a trig ratio.
1. A right-angled triangle has legs 5 and 12. Find the hypotenuse.
Answer: 13, because 5^2 + 12^2 = 25 + 144 = 169, and sqrt(169) = 13.
2. A right-angled triangle has a hypotenuse of 17 and one leg of 15. Find the other leg.
Answer: 8, because 17^2 - 15^2 = 289 - 225 = 64, and sqrt(64) = 8.
3. A right-angled triangle has legs 6 and 8 and hypotenuse 10. The angle theta is opposite the side of length 6. Find cos(theta).
Answer: 0.8, because cos(theta) = adjacent/hypotenuse = 8/10 = 0.8.
Teacher notes and timings
- Rough timing across four lessons: Lesson 1 finding the hypotenuse (section 1), Lesson 2 finding a missing leg (section 2), Lesson 3 trig ratios (section 3), Lesson 4 applying both to real-world problems plus the exit ticket (section 4 and assessment).
- This unit deliberately uses only whole-number Pythagorean triples (3-4-5, 5-12-13, 8-15-17 and their scaled multiples) so every answer is exact, matching this site's 'never wrong' answer-key policy; a calculator-based decimal unit can follow once the concept is secure.
- Language to repeat: the hypotenuse is always the longest side, opposite the right angle; SOH-CAH-TOA gives the three ratios; ratios depend on the ANGLE, not the triangle's size.
- Curriculum note: AC9M9M03 (Australian Curriculum v9) covers applying Pythagoras and trigonometry to spatial problems; AC9M9SP01 covers the similarity reasoning behind why the ratios are constant. This unit teaches both together since the application (section 4) depends on both.
- Present and print both work: use the Print button for a clean handout, or work the ladder problem live on the board, asking students to identify the hypotenuse before calculating anything.