Why trig ratios stay constant: a week of ready-to-teach maths

Five lessons for Year 9 Space. 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 12 triangle cards at actual size, a paper square and triangle to fold, a table of ratios, 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 at actual size (100%, no “fit to page”), one copy per pair: sheet 1 has the triangle cards (Days 3 and 4), sheet 2 the square, the triangle and the table of ratios (Days 4 and 5). Cut the cards before Day 3.
4. Each student needs a ruler marked in millimetres, a protractor and a calculator.
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

Two right-angled triangles with the same acute angle have all their angles equal, so they are similar: one is an enlargement of the other, and every side is multiplied by the same scale factor k. In a ratio of two sides the k is on the top and the bottom and cancels, so the ratio depends on the angle alone. That fixed ratio is what sine, cosine and tangent name: opposite over hypotenuse, adjacent over hypotenuse, and opposite over adjacent.

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 answers

Ratios of whole-number sides are fractions in lowest terms, then a decimal to three places with ≈ when it is rounded, as in 5/13 ≈ 0.385. Measured lengths are to the nearest millimetre, so a ratio from a measured triangle can differ from the table in its second or third decimal place. Exact values keep their surds, as in √3/2.

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 recalls similar triangles and scale factors; Day 2 meets a constant ratio in the real world, in shadows; Day 3 proves the ratio cannot change and tests it on printed triangles; Day 4 names the three ratios; and Day 5 finds exact values for the special angles. It builds on Year 8 work on enlargement and similar figures and on the trigonometric ratios met earlier in Year 9, and it prepares the enlargement transformations that come next in this strand.

Materials for the week (one trip)

• From the classroom: rulers marked in millimetres, protractors, calculators, scissors, envelopes, this pack printed.
• Cut-out sheet 1, one per pair, printed at actual size: 12 triangle cards (Days 3 and 4).
• Cut-out sheet 2, one per pair, printed at actual size: a 8 cm square and a 8 cm equilateral triangle to cut and fold (Day 5), and a table of ratios (Day 4).
• If the sun is out on Day 2: a metre ruler and a tape measure for each group. No other equipment to buy.

Words for the word wall

similar, enlargement, scale factor, corresponding sides, proportion, ratio, constant, opposite, adjacent, hypotenuse, sine, cosine, tangent, exact value, surd, rational, irrational

Dear families

This week in maths, Year 9 is finding out why the sides of a right-angled triangle keep the same ratios however big the triangle is drawn, as long as its angles stay the same. That fixed ratio is what the sine, cosine and tangent buttons on a calculator hold.

Try this at home

• On a sunny day, measure the shadow of something tall and your own shadow at the same time. Your height over your shadow should match its height over its shadow.
• Ask: a photo is enlarged so every length doubles. Do the angles in it change? (No: only the lengths.)
• Ask: a ramp rises 1 m for every 4 m along the ground. A longer ramp at the same angle runs 12 m along. How high does it rise? (3 m.)

My ratio week

Fill one row a day. Write a triangle or a pair of lengths you met, the ratio, and whether it stayed the same.

Similar triangles

Two triangles are similar when they have the same angles: the same shape, perhaps a different size. Then one is an enlargement of the other, and every side is multiplied by the same scale factor. Checking all three sides against one factor tells similar triangles apart from triangles that only look alike.

We are learning to

• decide whether two right-angled triangles are similar by comparing their sides,
• find the scale factor and use it to find missing sides,
• check that an enlargement keeps every angle the same.

Success criteria

• I can find a scale factor and test it on all three sides.
• I can draw an enlargement and show that its angles match.

You need

The worksheet; a ruler marked in millimetres, a protractor and a calculator for each student.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with the drawing.

Watch for these ideas

• Adding the same amount to every side instead of multiplying (3, 4, 5 becomes 5, 6, 7, which is not similar).
• Dividing the shortest side of one triangle by the longest of the other: list the sides shortest first and match them in order.
• Thinking an enlargement makes the angles bigger.

Answers

• Part A: 1 similar, k = 2; 2 similar, k = 1.25; 3 not similar (1.333, 1.875, 1.7); 4 similar, k = 0.5; 5 not similar (2.222, 1.75, 1.933).
• Part B: k = 20 ÷ 8 = 2.5, so 15, 20, 25; k = 87 ÷ 29 = 3, so 60, 63, 87.
• Part C: legs 6 cm and 8 cm, hypotenuse 10 cm; the acute angles are about 37° and 53° in both triangles.
• Part D: the angles of a triangle add to 180°, so with a right angle and one acute angle equal, the third angles are equal too; all three angles match, so the triangles are similar.

Similar triangles

A. Similar or not?

The sides of each pair of right-angled triangles are listed shortest first. Divide each side of the second by its match in the first.

B. Missing sides

C. Draw an enlargement

Draw a right-angled triangle with legs 3 cm and 4 cm. Then draw its enlargement by 2. Measure the two acute angles of each triangle with a protractor.

D. Stretch: one angle is enough

Two right-angled triangles each have a 35° angle. Explain why they must be similar.

Same angle, same ratio

At one moment the sun’s rays are parallel, so every upright object and its shadow make a right-angled triangle with the same angle at the tip of the shadow. The triangles are similar, so height over shadow is one number for all of them, and one measured object gives the height of anything else nearby.

We are learning to

• see that right-angled triangles with the same angle have the same side ratio,
• use that ratio to find a height or a shadow,
• draw a triangle from its ratio and measure its angle.

Success criteria

• I can find the height of a tree from its shadow and a post’s ratio.
• I can explain why the ratio is different later in the day.

You need

The worksheet; a protractor and a calculator each; if the sun is out, a metre ruler and a tape measure per group (Part B).

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second outside with Part B.

Watch for these ideas

• Subtracting (shadow minus height) instead of dividing.
• Height over shadow for one object and shadow over height for the next.
• Using shadows measured an hour apart, when the sun has moved.

Answers

• Part A: a fence post 1.2 ÷ 1.6 = 0.75; a student 1.5 ÷ 2 = 0.75; the tree is 0.75 × 6 = 4.5 m tall; the building’s shadow is 18 ÷ 0.75 = 24 m.
• Part B: results vary; at one time every object gives nearly the same height over shadow.
• Part C: a 2.4 cm by 3.2 cm right-angled triangle; the sun is about 37° above the ground.
• Part D: 1.2 ÷ 2.4 = 0.5, so the flagpole’s shadow is 7.5 ÷ 0.5 = 15 m. The sun is lower, so the angle and the ratio have changed, but at any one moment every object still shares one ratio.

Same angle, same ratio

A. Shadows at 10 am

All four shadows were measured at the same moment. Find each height over shadow, then the missing lengths.

B. Measure your own (on a sunny day)

Measure each height and shadow within a few minutes. Use the metre ruler’s ratio to find the height of something tall.

C. Draw the sun’s angle

Draw the fence post and its shadow as a right-angled triangle, with 1 cm for 0.5 m. Join the top of the post to the tip of the shadow, and measure the angle at the tip: the sun’s height above the ground.

D. Stretch: later in the day

At 3 pm the same 1.2 m post casts a 2.4 m shadow. How long is the shadow of a 7.5 m flagpole then? Has the rule “same angle, same ratio” broken?

Why the ratio stays

Enlarge a right-angled triangle by k and every side is multiplied by k. In any ratio of two sides, k sits on the top and the bottom and cancels, so the ratio is what it was. Any two right-angled triangles with the same acute angle are similar, so each ratio depends on the angle alone, and sorting triangles by a measured ratio sorts them by their angle.

We are learning to

• show that a scale factor cancels in every ratio of two sides,
• measure a triangle and work out the ratio for its marked angle,
• sort triangles of different sizes and positions by their ratio.

Success criteria

• I can explain why (k × opp) ÷ (k × hyp) = opp ÷ hyp.
• I can sort the twelve triangle cards into piles of equal ratio.

You need

The worksheet; the triangle cards (cut-out sheet 1) per pair; a ruler marked in millimetres and a calculator each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with the card sort.

Watch for these ideas

• Measuring the side next to the right angle as the opposite when the card is turned round: the opposite is the side facing the arc.
• Rounding lengths to the nearest centimetre, which blurs the piles; measure to the millimetre.
• Thinking a bigger card must have a bigger ratio.

Answers

• Part A: for the angle opposite the 5, every row gives opp ÷ hyp = 5/13 ≈ 0.385, adj ÷ hyp = 12/13 ≈ 0.923 and opp ÷ adj = 5/12 ≈ 0.417; the sides are 10-24-26, 15-36-39, 2.5-6-6.5, 50-120-130.
• Part B: k × 5, k × 12 and k × 13; (k × 5) ÷ (k × 13) = 5/13, because k ÷ k = 1.
• Part C: the card key is with the mini-check answers.
• Part D: no. A different ratio means a different angle, so the two triangles are different shapes and cannot be similar.

Why the ratio stays

A. One family

The 5-12-13 triangle enlarged by each k. For the angle opposite the 5, write each ratio as a fraction and a decimal.

B. Why it works

An enlargement by k has opposite ________, adjacent ________ and hypotenuse ________. Its opposite over hypotenuse is ____________ = ________, because ______________________________.

C. Sort the triangle cards

For the marked angle of each card, measure the opposite and adjacent sides to the millimetre and divide. Then sort the cards into piles of equal ratio.

D. Stretch: similar or not?

One right-angled triangle has opp ÷ adj = 0.75 for its marked angle; another has 0.8. Can they be similar? Explain.

Sine, cosine and tangent

Each constant ratio has a name. For a marked angle θ: sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. The opposite and the adjacent depend on which angle is marked, not on how the triangle is turned; the hypotenuse always faces the right angle. A calculator or a table can hold the values because each angle has only one of each.

We are learning to

• name the opposite, adjacent and hypotenuse for a marked angle in any position,
• write the sine, cosine and tangent of both acute angles as fractions,
• find an angle from a measured ratio in a table.

Success criteria

• I can label the three sides of a turned triangle for its marked angle.
• I can find the angle of each card pile from the table.

You need

The worksheet; the card piles from Day 3; the table of ratios (cut-out sheet 2); a ruler and a calculator each.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part B; start the second with the card piles.

Watch for these ideas

• Calling the side next to the right angle “adjacent” whichever angle is marked.
• Swapping sine and cosine.
• Taking the right angle as the marked angle.

Answers

• Part A: 1 opposite BC, adjacent AC, hypotenuse AB; 2 opposite QR, adjacent PR, hypotenuse PQ; 3 opposite YZ, adjacent XZ, hypotenuse XY; 4 opposite MN, adjacent KN, hypotenuse KM.
• Part B: opposite 8: sin 8/17, cos 15/17, tan 8/15; opposite 15: sin 15/17, cos 8/17, tan 15/8; opposite 20: sin 20/29, cos 21/29, tan 20/21; opposite 21: sin 21/29, cos 20/29, tan 21/20. The sine of one acute angle is the cosine of the other.
• Part C: the three piles have tangents near 0.466, 0.839, 1.428: angles of 25°, 40°, 55°.
• Part D: the two acute angles add to 90°, and the side opposite one is adjacent to the other, so sin A = cos(90° − A).

Sine, cosine and tangent

A. Label the sides

For the marked angle θ in each triangle, name the opposite, adjacent and hypotenuse by their corners.

B. Both acute angles

C. Find the angles

Take one card from each Day 3 pile. Use its opp ÷ adj and the table of ratios to find its angle, then check with opp ÷ hyp.

D. Stretch: sine and cosine swap

Explain why the sine of one acute angle of a right-angled triangle is always the cosine of the other.

Exact values

Two folds give exact values. A square folded on its diagonal makes a right-angled triangle with two 45° angles and equal legs, so tan 45° = 1, and with the hypotenuse s√2, sin 45° = cos 45° = √2/2. An equilateral triangle folded in half makes a triangle with angles of 30°, 60° and 90° and sides s/2, s√3/2 and s. Values with a surd are irrational: a decimal only approximates them.

We are learning to

• find the exact sine, cosine and tangent of 30°, 45° and 60° from two folded shapes,
• tell rational values from irrational ones,
• explain why the size of the folded shape does not matter.

Success criteria

• I can find sin 30° and cos 30° from the folded triangle as exact values.
• I can say which of the nine values are rational.

You need

The worksheet; the square and the equilateral triangle (cut-out sheet 2), scissors and a ruler per pair; the unit’s self-check quiz.

Lesson flow (about 60 minutes)

Two shorter sessions instead? Split after Part A; start the second with the triangle.

Watch for these ideas

• Writing sin 45° = 0.707 with an equals sign.
• Measuring the fold of the triangle and calling 6.9 its exact length.
• Thinking a bigger square would give bigger values.

Answers

• Part A: legs 8 cm, diagonal √128 = 8√2 ≈ 11.3 cm; sin 45° = cos 45° = 8/(8√2) = √2/2 ≈ 0.707; tan 45° = 8/8 = 1.
• Part B: sides 4 cm, 8 cm and √(64 − 16) = √48 = 4√3 ≈ 6.9 cm; sin 30° = 1/2, cos 30° = √3/2, tan 30° = √3/3, sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.
• Part C: rational: sin 30° = 1/2, tan 45° = 1, cos 60° = 1/2; the other six are irrational.
• Part D: a square of side s has diagonal s√2, so sin 45° = s/(s√2): the s cancels and leaves √2/2 whatever the size.

Exact values

A. The folded square

Cut out the square and fold it along its diagonal. Each half is a right-angled triangle with two 45° angles.

B. The folded triangle

Cut out the equilateral triangle and fold one corner onto another. Each half has angles of 30°, 60° and 90°.

C. The nine exact values

Fill the table from Parts A and B, then circle the values that are rational.

D. Stretch: any size

Explain why a square of any size gives the same value for sin 45°.

Twelve triangle cards

Days 3 and 4, one set per pair. Print at actual size. Each card is a right-angled triangle; the arc marks the angle to work with. Measure to the millimetre.

Teacher note: the key is with the mini-check answers, so this sheet can go to pairs whole.

A square, a triangle and a table of ratios

Print at actual size. Day 5: cut out the square and the equilateral triangle along the dashed edges and fold each along its fold line. Day 4: use the table to find the angles of the card piles.

Sine, cosine and tangent, to three decimal places

What we know: Why trig ratios stay constant

Work on your own. Give ratios as fractions, and as decimals where asked. Show your working.

1. Are right-angled triangles with sides 12, 35, 37 and 36, 105, 111 similar? If so, give the scale factor. ______________________
2. A 9-40-41 triangle is enlarged by 5. What is the sine of the angle opposite the side of 9 in the enlargement? ______________________
3. At one moment a 2 m pole casts a 2.5 m shadow. How tall is a tree whose shadow is 15 m? ______________________
4. A right-angled triangle has opposite 7 and hypotenuse 25 for its angle θ. Find the adjacent side, sin θ and cos θ. ______________________
5. Find tan θ for the triangle in Question 4. ______________________
6. Two students draw a 50° right-angled triangle at different sizes and measure opp ÷ adj as 1.19 and 1.2. Why do they differ? ______________________
7. Write sin 60° exactly. ______________________
8. Is tan 60° rational or irrational? Give its exact value. ______________________

Answers and marking guide

Answers with working

1. Yes: 36 ÷ 12, 105 ÷ 35, 111 ÷ 37 all equal 3, so k = 3.
2. (5 × 9) ÷ (5 × 41) = 9/41 ≈ 0.22: the 5 cancels.
3. 2 ÷ 2.5 = 0.8, so 0.8 × 15 = 12 m.
4. √(625 − 49) = 24; sin θ = 7/25, cos θ = 24/25.
5. 7/24 ≈ 0.292.
6. Measuring error: both triangles have the same angle, so the true ratio is the same, tan 50° ≈ 1.192.
7. √3/2.
8. Irrational: tan 60° = √3.

A quick three-level guide

Eight questions, four ideas. A student at standard finds ratios from sides and explains that the angle alone fixes them.

Card key

Triangle cards (Days 3 and 4): 25°, opp ÷ adj ≈ 0.47: A, D, F, K, L; 40°, opp ÷ adj ≈ 0.84: C, E, H, J; 55°, opp ÷ adj ≈ 1.43: B, G, I. Measured to the millimetre, a card can be 0.02 or so off.

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: similar triangles, same angle and same ratio, why k cancels, naming the ratios, and exact values.