Pythagoras and trigonometry: a week of ready-to-teach maths

Five lessons for Year 10 Measurement. Print this pack and the week is prepared: each day has a one-page plan and a student worksheet with practice and a stretch section, plus twelve ratio cards to match, twelve problem cards to sort, a mini-check and every answer with its working.

Start here: five minutes to Monday

1. Skim the week at a glance on the next page.
2. Print the five days. Each day is two A4 sheets: a plan and a worksheet.
3. Print the two cut-out sheets, one copy per pair. Cut the 12 ratio cards before Day 2 and the 12 problem cards before Day 4, and keep each set in an envelope.
4. Calculators are used every day, set to degrees. Rulers and protractors help on Day 2.
5. Open the free interactive unit on the board. Every plan says which picture to show and which buttons to press.

The idea in plain words

Right-angled triangles turn up everywhere: in ladders and ramps, in a tree and its shadow, in a ship’s course. Pythagoras’ theorem links the three sides; trigonometry links the sides to the angles. The three ratios, sine, cosine and tangent, depend only on the angle, not on the size of the triangle, which is why one angle and one length are enough to find everything else. Angles of elevation and depression are measured from the horizontal, and bearings from north, turning clockwise.

One day, one lesson

The five lessons fill a week of maths, one lesson of about 60 minutes a day. Run them in order: each day stands on the one before. Every plan marks a split point for two shorter sessions, and every worksheet ends with a stretch section for students who finish early.

How we write these

Sides are named from the angle θ: the opposite, the adjacent, and the hypotenuse, opposite the right angle. Lengths are given to two decimal places and angles to one, such as 33.7°. True bearings have three figures, such as 040°.

The week at a glance

One lesson a day for a week. Each day stands on the day before, so run them in order.

How the week builds

Day 1 works with the sides alone; Day 2 introduces the ratios that link sides and angles; Days 3 and 4 use them to find sides and angles in practical problems; and Day 5 turns them to direction. It builds on Year 9, where students first met Pythagoras’ theorem and trigonometry in right-angled triangles, and it prepares the sine and cosine rules of senior mathematics.

Materials for the week (one trip)

• From the classroom: calculators, rulers, protractors, scissors, envelopes for the cards, this pack printed.
• Cut-out sheet 1, one per pair: 12 ratio cards (Day 2).
• Cut-out sheet 2, one per pair: 12 problem cards (Day 4).
• If you have them: a clinometer, or a protractor with a straw and a weight on a string, to measure an angle of elevation outside on Day 4. No other equipment to buy.

Words for the word wall

hypotenuse, opposite, adjacent, Pythagoras’ theorem, sine, cosine, tangent, inverse, angle of elevation, angle of depression, horizontal, bearing, true bearing, compass direction, clinometer

Dear families

This week in maths, Year 10 is solving practical problems with right-angled triangles: how far up a wall a ladder reaches, how tall a tree is from its distance and an angle, and how far north a ship has sailed on a bearing.

Try this at home

• Ask: a ladder 5 m long stands 3 m out from a wall. How high does it reach? (4 m.)
• Ask: why do builders check a corner with 3 m, 4 m and 5 m? (Those three lengths only fit around a right angle.)
• Ask: what bearing is due west? (270°.)
• Ask: looking up at a kite, is that an angle of elevation or of depression? (Elevation.)

My right angles week

Fill one row a day. Write a right angle you noticed, what you could work out from it, and how.

Pythagoras in practice

Pythagoras’ theorem links the three sides of a right-angled triangle. Students find hypotenuses and shorter sides, solve ladder, screen and ramp problems, check whether a corner is square, and find the longest rod that fits in a box.

We are learning to

• use c² = a² + b² to find the hypotenuse, or a shorter side,
• draw and label a right-angled triangle from a practical situation,
• use the theorem in reverse to test whether an angle is a right angle.

Success criteria

• I can find how high a 6 m ladder reaches when its foot is 1.5 m from the wall.
• I can explain why a corner with sides 3 m, 4 m and 5.1 m is not square.

You need

Calculators; the worksheet, one per student.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Adding the squares to find a shorter side: subtract from the square of the hypotenuse.
• Forgetting the square root at the end: c² = 100 means c = 10.
• Taking any side as the hypotenuse: it is always the longest side, opposite the right angle.

Answers

• Part A: hypotenuses 10.00, 13.00, 11.40; shorter sides 24.00, 8.00, 12.00.
• Part B: ladder √(6² − 1.5²) ≈ 5.81 m up; screen diagonal ≈ 137.93 cm; ramp ≈ 6.02 m long.
• Part C: 3² + 4² = 25, but 5.1² = 26.01. They are not equal, so the corner is not square; the long side is too long, so the angle is a little more than 90°.
• Part D: floor diagonal √(30² + 40²) = 50 cm, then √(50² + 120²) = 130 cm.

Pythagoras in practice

A. Find the missing side

Give answers to two decimal places where needed. The first is done for you.

B. Real right angles

Draw and label a right-angled triangle for each problem first.

C. Is it square?

D. Stretch: the longest rod

Three ratios

For an angle θ in a right-angled triangle the sides are named opposite, adjacent and hypotenuse, and the three ratios compare them. Students name the sides from both acute angles of one triangle, see that a ratio stays fixed as the triangle grows, and match six triangles to their ratios.

We are learning to

• name the opposite, adjacent and hypotenuse for a given angle,
• write sin θ, cos θ and tan θ as fractions of sides,
• explain why each ratio depends only on the angle, not on the size of the triangle.

Success criteria

• I can write the three ratios for a 5-12-13 triangle from either acute angle.
• I can explain why every right-angled triangle with a 30° angle has sin 30° = 0.5.

You need

Calculators; rulers and protractors; the ratio cards (cut-out sheet 1), one set per pair; the worksheet.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Naming the sides without looking at θ: opposite and adjacent swap when the angle changes.
• “tan uses the hypotenuse.” Tangent is opposite over adjacent.
• “A bigger triangle has a bigger sine.” The ratio depends only on the angle.

Answers

• Part A: from θ, sin θ = 5/13, cos θ = 12/13, tan θ = 5/12; from the other angle, sin φ = 12/13, cos φ = 5/13, tan φ = 12/5.
• Part B: opposite sides about 2, 4, 6 cm, so every ratio is 0.5; the calculator gives sin 30° = 0.5.
• Part C: the card key is with the mini-check answers.
• Part D: the hypotenuse is the longest side, so opposite ÷ hypotenuse and adjacent ÷ hypotenuse are at most 1. For the 3-4-5 triangle (3/5)² + (4/5)² = 9/25 + 16/25 = 1: Pythagoras divided by the hypotenuse squared.

Three ratios

A. Name the sides, write the ratios

B. Same angle, different sizes

Draw three right-angled triangles with a 30° angle and hypotenuses of 4, 8, 12 cm. Measure each opposite side.

C. Match the triangles

Match each triangle card to the ratio card that fits its marked angle. Write the letters.

D. Stretch: never more than 1

Finding a side

With one side and one angle, a ratio gives any other side. Students choose the ratio that links the side they know with the side they want, and find a kite’s height, a ramp’s rise and a tree’s height.

We are learning to

• choose sin, cos or tan from the side we know and the side we want,
• rearrange to find a side, including when it is on the bottom of the fraction,
• solve practical problems, adding an eye height where one is needed.

Success criteria

• I can find x when the hypotenuse is 10 and the angle 40°, with x opposite the angle.
• I can find the height of a tree from 25 m away at an angle of elevation of 38°.

You need

Calculators, set to degrees; the worksheet, one per student.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• A calculator in radians: check that sin 30 gives 0.5.
• Dividing when x is on top: if sin θ = x/10, then x = 10 sin θ.
• Leaving out the eye height when an angle is measured from where someone stands.

Answers

• Part A: 1. sin, x = 6.43; 2. cos, x = 7.25; 3. tan, x = 17.88; 4. sin, x = 10.19.
• Part B: kite 60 sin 55° ≈ 49.15 m; ramp 4 sin 6° ≈ 0.42 m; tree 25 tan 38° ≈ 19.53 m, plus 1.6 m of eye height, about 21.13 m.
• Part C: knowing the adjacent side and wanting the opposite side: tan; knowing the hypotenuse and wanting the adjacent side: cos; knowing the opposite side and wanting the hypotenuse: sin.
• Part D: tan⁻¹(1/14) ≈ 4.1°. To rise 0.6 m it needs a run of 8.4 m, and the ramp is ≈ 8.42 m long.

Finding a side

A. Find x

Name the sides from the marked angle, choose the ratio, and find x to two decimal places. The first is done for you.

B. Sides in practice

C. Choose and justify

D. Stretch: one in fourteen

Finding an angle

Inverse ratios turn a ratio back into an angle. Students find angles from two sides, work with angles of elevation and depression, and sort twelve problems by the tool that solves them.

We are learning to

• use sin⁻¹, cos⁻¹ and tan⁻¹ to find an angle from two sides,
• draw angles of elevation and depression from the horizontal,
• choose between Pythagoras, a ratio and an inverse ratio for a problem.

Success criteria

• I can find the angle when the opposite side is 11 and the adjacent side 4.
• I can explain why the angle of depression from a lighthouse equals the angle of elevation from the boat.

You need

Calculators, set to degrees; the problem cards (cut-out sheet 2), one set per pair; a clinometer if you have one; the worksheet.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; start the next with Part B.

Watch for these ideas

• Measuring from the vertical: elevation and depression are both measured from the horizontal.
• Pressing tan instead of tan⁻¹: the answer comes out as a ratio, not an angle.
• Using a pair of sides that does not match the inverse: sin⁻¹ needs the opposite and the hypotenuse.

Answers

• Part A: 1. sin⁻¹(5/9) ≈ 33.7°; 2. cos⁻¹(7/10) ≈ 45.6°; 3. tan⁻¹(11/4) ≈ 70.0°.
• Part B: 1. tan⁻¹(45/120) ≈ 20.6°; the angle of elevation from the boat is the same, as alternate angles between the two horizontal lines. 2. 80 ÷ tan 30° ≈ 138.56 m.
• Part C: the card key is with the mini-check answers.
• Part D: with h the height, h ÷ tan 30° − h ÷ tan 40° = 20, so h ≈ 37.02 m.

Finding an angle

A. Find θ

Choose the inverse ratio that uses the two sides given, and find θ to one decimal place. The first is done for you.

B. Up and down

Draw a diagram for each, with the horizontal line marked.

C. Which tool?

Sort the twelve problem cards by the tool that solves each one, then write the letters. Solve one card from each pile.

D. Stretch: two sightings

Bearings

A true bearing gives a direction as the angle turned clockwise from north, in three figures. Students write compass directions as bearings, split a ship’s course into north and east parts, find a distance and a bearing from two parts, and play the self-check quiz.

We are learning to

• write a direction as a three-figure true bearing,
• use sin and cos to find how far east and how far north a course goes,
• find the distance and the bearing of one place from another.

Success criteria

• I can write south-west as a bearing of 225°.
• I can find how far north a ship goes on a bearing of 040° for 30 km.

You need

Calculators, set to degrees; protractors; the worksheet, one per student; the self-check quiz on the board.

Lesson flow (about 60 minutes)

Two shorter sessions? Stop after Part A; do Parts B to D and the quiz next time.

Watch for these ideas

• Measuring anticlockwise, or from east: bearings turn clockwise from north.
• Two-figure bearings: 40° is written 040°.
• Swapping sin and cos: with the angle at the north line, the north part is adjacent (cos) and the east part opposite (sin).

Answers

• Part A: North = 000°; East = 090°; South-west = 225°; North-west = 315°; S 30° E = 150°; N 20° W = 340°.
• Part B: east 30 sin 40° ≈ 19.28 km; north 30 cos 40° ≈ 22.98 km; the way back is on 220°.
• Part C: distance √(6² + 8²) = 10 km; bearing tan⁻¹(6/8) ≈ 36.9°, so 037°.
• Part D: 14.14 km from the start, on a bearing of 135°.

Bearings

A. Directions as bearings

Write each direction as a three-figure true bearing. The first is done for you.

B. A ship’s course

C. Distance and bearing

D. Stretch: two legs

Triangles and their ratios

For Day 2, one set per pair. Six cards show a right-angled triangle with an angle θ marked; six give one ratio of θ. Match each triangle to its ratio. Name the sides from θ first.

Teacher note: the key is with the mini-check answers.

Which tool solves it?

For Day 4, one set per pair. Sort the cards by the tool each problem needs: Pythagoras, a ratio to find a side, or an inverse ratio to find an angle. Then solve one card from each pile.

Teacher note: the key is with the mini-check answers.

What we know: Pythagoras and trigonometry

Work on your own. A calculator in degrees is allowed. Draw a diagram for each problem; give lengths to two decimal places.

1. A 5 m ladder reaches 4.5 m up a wall. How far is its foot from the wall? ______________________
2. Write tan θ and cos θ for this triangle. ______________________
3. The hypotenuse is 12 cm and the angle 35°. Find the side opposite the angle. ______________________
4. The opposite side is 5 and the adjacent side 9. Find the angle. ______________________
5. From 40 m away, the angle of elevation of the top of a building is 52°. How tall is it? ______________________
6. An angle of depression is measured, circle one:   up from the horizontal  ·  down from the horizontal  ·  clockwise from north
7. A boat sails 20 km on a bearing of 130°. How far east and how far south has it gone? ______________________
8. From the top of a 30 m cliff, a boat is at an angle of depression of 25°. How far is the boat from the foot of the cliff? ______________________

Answers and marking guide

Answers with working

1. √(5² − 4.5²) = √4.75 ≈ 2.18 m.
2. tan θ = 7/24, cos θ = 24/25.
3. 12 sin 35° ≈ 6.88 cm.
4. tan⁻¹(5/9) ≈ 29.1°.
5. 40 tan 52° ≈ 51.20 m (plus eye height if it was measured standing).
6. down from the horizontal.
7. East 20 sin 130° ≈ 15.32 km; south 20 cos 50° ≈ 12.86 km.
8. 30 ÷ tan 25° ≈ 64.34 m.

A quick three-level guide

Eight questions, four ideas. A student at standard draws a labelled diagram for each problem and answers with units.

Card keys

Ratio cards (Day 2): G and A (sin θ = 3/5), D and J (tan θ = 4/3), E and L (cos θ = 12/13), B and H (sin θ = 12/13), K and F (tan θ = 8/15), I and C (cos θ = 8/17). Problem cards (Day 4): pythagoras: M, R, W, X; a ratio, for a side: N, P, Q, S, V; an inverse ratio, for an angle: O, T, U.

Weekly class record

Jot a tick as you move around the room; the mini-check fills any gaps. A tick a day is plenty.

The five columns are the five days: Pythagoras, the ratios, finding sides, finding angles, and bearings.