A side on top
θ = 35°, hypotenuse 10, find the opposite side x
O and H: sin 35° = x/10
x = 10 × sin 35° = 5.735… = 5.7 (1 decimal place)
Trigonometry in right-angled triangles Why does one angle always give the same ratio of sides, however big the triangle?
Foundation and Higher · AQA · 45 minutes
Students find that triangles of one shape on a grid share a ratio of sides, then label the sides from the angle, choose sin, cos or tan, find sides and angles, and build the exact values from two triangles. Slides to present, a vote that shows who's stuck, practice and homework.
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A circle of radius 1. The angle at (0, 0) is 15n degrees: move n from 0 to 6, from 0° to 90°. The point on the circle is (cos θ, sin θ). Read it at 30° (n = 2) and at 60° (n = 4). What do you notice?
They cross at (0.87, 0.5)
The idea in one minute
Right-angled triangles with the same angle are the same shape, so the ratio of any two of their sides is the same, however big they are. Those ratios have names: sin θ = opposite ÷ hypotenuse, cos θ = adjacent ÷ hypotenuse, tan θ = opposite ÷ adjacent. Name the sides from the angle you are using, choose the ratio that has the two sides in the question, and solve: a side from the ratio, an angle from the inverse (sin⁻¹, cos⁻¹, tan⁻¹). On a circle of radius 1, the point at angle θ is (cos θ, sin θ).
The lesson goes picture first. Students find up ÷ across for three triangles of one shape on the grid before anyone says tan. Then one habit for every question: label H, O and A from the angle, choose the ratio, then write the equation. Exact values for 0°, 30°, 45°, 60° and 90° come from half a square and half an equilateral triangle. The core is for Foundation and Higher.
By the end, students can
- label the hypotenuse, opposite and adjacent sides from a given angle
- choose sin, cos or tan from the two sides a question has
- find a missing side, including one under the line, such as tan 40° = 6/x
- find a missing angle with sin⁻¹, cos⁻¹ or tan⁻¹, in degrees
- give exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°
Key words
- hypotenuse
- opposite
- adjacent
- sine
- cosine
- tangent
- inverse
- exact value
Before this lesson
Students should know Pythagoras' theorem, find the hypotenuse, solve x/5 = 3 and 12/x = 4, and round to 1 decimal place. Trigonometry is taught here.
Taught in Pythagoras' theorem.
The lesson, step by step
60 minutes in total. Each step says what changes for this length.
6 min 1 · Hook · students think
Same shape, same ratio
Three triangles of one shape on the grid: students find up ÷ across for each before anyone says tan.
Say“Three right-angled triangles, all with the same angle at (0, 0). For each one, work out up ÷ across. Don't show anyone yet.”
Ask- What is up ÷ across for each triangle?
- Draw a fourth triangle with the same angle. What will up ÷ across be?
Watch forWorking out across ÷ up (2, not 0.5): ask which they were asked for, and whether it matters, as long as every triangle is done the same way.
For the teacher
The triangles go 2 across and 1 up, 4 across and 2 up, 6 across and 3 up: each is the last one enlarged, so the angle at (0, 0) stays the same, and up ÷ across is 0.5 every time.
Don't name it yet. The question is the point: if the angle fixes the ratio, can the ratio find a side, or the angle?
Gather answers with Ask the class: 0.5 every time. The angle is about 26.6°; keep it for when tan⁻¹ comes.
Shorter: Show the three ratios at once; skip the working out (−2 min).
13 min 2 · Explain · you explain
Label, choose, solve
Name the sides from the angle; SOHCAHTOA; a side, a side under the line, an angle; then exact values.
Say“Every question goes the same way: label the sides from the angle, choose the ratio with the two sides you have, then solve.”
Ask- If the angle moves to the other corner, which side is opposite now?
- Before the calculator: will this angle be more or less than 45°?
Watch forLabelling the hypotenuse as the adjacent: it touches the angle too, but it is always the hypotenuse.
A side under the line
θ = 40°, opposite 6, find the adjacent side x
O and A: tan 40° = 6/x
x = 6 ÷ tan 40° = 7.150… = 7.2 (1 decimal place)
An angle
Opposite 5, hypotenuse 8, find θ
O and H: sin θ = 5/8
θ = sin⁻¹(5/8) = 38.7° (1 decimal place)
An exact value
Half an equilateral triangle of side 2: hypotenuse 2, short side 1, height √3
sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 = √3/3
For the teacher
First slide: the hypotenuse is opposite the right angle; the opposite is opposite the angle θ; the adjacent is next to θ and isn't the hypotenuse. sin θ = O/H, cos θ = A/H, tan θ = O/A: SOH CAH TOA. In the 3, 4, 5 triangle with θ at (0, 0), sin θ = 3/5 and tan θ = 3/4.
Second slide: on a circle of radius 1 the hypotenuse is 1, so sin θ is the point's height and cos θ its distance across. Move n: at 30° the point is (0.87, 0.5), at 60° it is (0.5, 0.87). The two swap, because the triangle's other angle is 90° − θ.
Third slide: a side. When x is on top, multiply: x = 10 × sin 35°. When x is under the line, divide: x = 6 ÷ tan 40°. Ask: is the answer sensible? The hypotenuse is always the longest.
Fourth slide: an angle. The ratio is known and the angle is wanted, so use the inverse: θ = sin⁻¹(5/8). Keep the fraction as it is until the end. A calculator in radians gives 0.68 here: sin 90 must give 1.
Fifth slide: exact values. Half a square has sides 1, 1 and √2: so sin 45° = cos 45° = 1/√2 = √2/2, and tan 45° = 1. Half an equilateral triangle of side 2 has sides 1, √3 and 2: sin 30° = 1/2, cos 30° = √3/2, tan 30° = √3/3, and the same sides give 60° the other way round. At 0° and 90° the triangle flattens: sin 0° = 0, cos 0° = 1, sin 90° = 1, cos 90° = 0; tan 90° has no value.
Vote before each calculation (Ask the class): sin, cos or tan? Then: more or less than 45°? Have every calculator show sin 90 = 1 before the first angle.
Shorter: One side example, the angle, and exact values as the table only (−5 min).
8 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
Say“Watch me do one. Then you do its twin, which is almost the same. Label H, O and A first, every time.”
Ask- Which two sides does the question have? So which ratio?
Watch forMultiplying when x is under the line: tan 50° = 7/x gives x = 7 ÷ tan 50°.
A side on top
θ = 30°, hypotenuse 12: the opposite x?
sin 30° = x/12
x = 12 × 0.5 = 6
Then theirs
θ = 60°, hypotenuse 9: the adjacent x?
A side under the line
θ = 25°, opposite 4: the hypotenuse x?
sin 25° = 4/x
x = 4 ÷ sin 25° = 9.464… = 9.5 (1 dp)
Then theirs
θ = 50°, opposite 7: the adjacent x, to 1 dp?
An angle
Opposite 5, adjacent 12: θ?
tan θ = 5/12
θ = tan⁻¹(5/12) = 22.6° (1 dp)
Then theirs
Adjacent 7, hypotenuse 10: θ, to 1 dp?
For the teacher
Work the first example aloud, writing H, O and A on a sketch first. Then students solve its twin while you watch who labels before choosing a ratio.
Do the same for each pair. The second pair catches students who multiply when x is under the line; the third, students who use tan instead of tan⁻¹.
Open the answers slide once most have finished their twins.
Shorter: Do the first and third pairs only (−3 min).
23 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels, and a stretch.
For this length: The stretch, then: why does sin 30° equal cos 60°?
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- From angle A, which side is opposite? Which is the hypotenuse?
Watch forLabels taken from the wrong corner, or the side next to the angle taken for the hypotenuse.
Label H, O and A on a sketch first. Keep your calculator in degrees. Round only at the end, as each question says. The stretch is without a calculator.
The question that shows who's stuck
- Triangle ABC has its right angle at B. Angle A is 35° and AB = 8 cm. Which gives the length of BC?
In your Classroom, as they answer (an example) - A7
- B6
- C10
- D5
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- Triangle PQR has its right angle at Q. First the angle θ is at P, then at R. Sort each side into what it is from that angle.They sort 6 items into Hypotenuse and Opposite θ and Adjacent to θ.
- Which ratio would you use? Sort each question by the two sides it has.They sort 6 items into sin and cos and tan.
- A right-angled triangle has an angle of 30° and hypotenuse 14 cm. Find the side opposite the 30° angle, in cm.
- A right-angled triangle has an angle of 52°. The side next to it (not the hypotenuse) is 9 cm. Find the side opposite the 52° angle, in cm to 1 decimal place.
- The opposite side is 8 cm and sin θ = 0.4. So 8/x = 0.4, where x is the hypotenuse. Solve it line by line.They solve 8/x = 0.4 line by line, each line checked as they write it.
- A right-angled triangle has hypotenuse 12 cm. The side opposite the angle θ is 7 cm. Find θ in degrees, to 1 decimal place.
- Sam types sin⁻¹(0.5) into a calculator and gets 0.52. The angle should be 30°. What went wrong?
- Without a calculator: mark cos 30° on the number line.They drag a marker on a number line from 0 to 1.
- A ladder 6 m long leans against a wall and reaches 5.2 m up it. Find the angle between the ladder and the flat ground, to the nearest degree.
- Stretch, without a calculator: a right-angled triangle has hypotenuse 10 cm and an angle of 60°. The side next to the 60° angle is ___ cm, and the side opposite it is exactly ___ cm. Write the second as a surd: type √ or sqrt, such as 2√5 or 2sqrt5.
Support · For question 1: put your finger on the right angle; the side across from it is the hypotenuse. Put your finger on θ; the side across from it is the opposite. The side left is the adjacent. Then choose: O and H, sin; A and H, cos; O and A, tan.
For the teacher
Everyone chooses before anyone explains. Sketch triangle ABC with the right angle at B and ask the class to label H, O and A from angle A, then choose the ratio.
Questions 1–2 sort sides and ratios; 3–4 find sides; 5 solves an equation with x under the line, line by line; 6 finds an angle; 7 asks for a reason; 8 places an exact value; 9 is a problem in words. The stretch is exact, without a calculator.
Students show their working on paper or the board; the quiz keeps their final answers.
Ask two students who chose different answers to label the triangle on the board, before you say which is right.
Shorter: After the vote, questions 1–6 only (−5 min).
Longer: The stretch, then: why does sin 30° equal cos 60°? (+10 min).
10 min 5 · Exit · students show
One side, three exact values
A side found and three exact values show who labels and chooses the ratio, and who knows the special angles.
For this length: Make up a question whose answer is exactly 45°
Say“Two questions. Label the triangle before you start the first.”
Watch forUsing sin for the side next to the angle.
Label the triangle first. One sentence for the second part.
- A right-angled triangle has hypotenuse 15 cm and an angle of 40°. Find the side next to the 40° angle (not the hypotenuse), in cm to 1 decimal place.
- Without a calculator, write down sin 30°, cos 60° and tan 45°. Then say in one sentence where one of them comes from.
For the teacher
Find a side next to a given angle from the hypotenuse.
Then write three exact values, and say where one of them comes from.
The labels show who names sides from the angle: start the next lesson with a triangle turned on its side if few do. Set the homework.
Longer: Make up a question whose answer is exactly 45° (+5 min).
One to one: for every question, have the student write H, O and A on the triangle before anything else, and say aloud which two sides the question has; check angle answers by putting them back into the ratio.
Homework: sides, angles and exact values
Sketch each triangle and label H, O and A from the angle before you choose a ratio. Keep your calculator in degrees. Questions 1 and 6 are without a calculator.
- A right-angled triangle has an angle of 30° and hypotenuse 18 cm. Find the side opposite the 30° angle.
- A right-angled triangle has an angle of 48°. The side next to it (not the hypotenuse) is 7 cm. Find the side opposite it, to 1 decimal place.
- A right-angled triangle has an angle of 65°. The side opposite it is 12 cm. Find the hypotenuse, to 1 decimal place.
- The two sides that meet at the right angle are 4 cm and 9 cm. Find the angle opposite the 4 cm side, to 1 decimal place.
- A ramp 12 m long rises 1.5 m. Find the angle between the ramp and the flat ground, to 1 decimal place.
- Find the exact value of sin 60° × cos 30° + tan 45°.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching trigonometry in right-angled triangles: common questions
How do you know which side is opposite and which is adjacent?
Find the hypotenuse first: it is opposite the right angle. Then look at the angle you are using: the side across from it is the opposite, and the side left over, touching it, is the adjacent. Move to the other angle and the opposite and adjacent swap; the hypotenuse stays.
Why does SOHCAHTOA work for any size of triangle?
Right-angled triangles with the same angle are the same shape: one is an enlargement of the other, so every side grows by the same scale factor and the ratio of two sides doesn't change. That ratio belongs to the angle, and sin, cos and tan are its names.
My student's calculator gives strange answers. Why?
It is most likely set to radians or gradians. sin 90 should give 1; if it gives 0.89 (radians) or 0.99 (gradians), set it to degrees, with a D on the screen. An angle answer below 2 for a triangle is a warning sign. GCSE uses degrees throughout.
Do students need to learn the exact values by heart?
It helps, but they can rebuild them in a minute from half a square (sides 1, 1, √2) and half an equilateral triangle of side 2 (sides 1, √3, 2). sin read from 0° to 90° is cos read from 90° to 0°, which halves what there is to remember. tan 90° has no value.
Is this trigonometry lesson suitable for Foundation GCSE?
Yes. Labelling, finding sides and angles, and the exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90° are for Foundation and Higher. The sine rule, the cosine rule and trigonometry in three dimensions are later, Higher, lessons.
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Curriculum reference: AQA GCSE Mathematics 8300: G20 (the trigonometric ratios sin, cos and tan, to find angles and lengths in right-angled triangles) and G21 (exact values of sin θ and cos θ for 0°, 30°, 45°, 60° and 90°, and of tan θ for 0°, 30°, 45° and 60°).
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