On the Foundations Check, angles, shape and Pythagoras came back secure to step 5 and broke at step 6. Step 5 is Core standard; step 6 is the Extended rung. So Pythagoras itself is not a mystery to you — what broke is the layer where the triangle is not handed to you neatly, where you have to decide whether you are after the longest side or a shorter one, and where the answer is meant to stay as a surd.
That is why section 1 starts below your break, at the picture, rather than at the formula. You already know
a² + b² = c². Knowing it is not the problem. Knowing which letter is which in this
particular triangle is.
Two other findings from the check land squarely here:
−2ab cos C term of the cosine rule, where a negative cosine turns the whole term
positive and half of all students subtract anyway. Section 5 attacks that directly.| Code | Sub-topic | Expected difficulty for you |
|---|---|---|
| E6.1 | Pythagoras’ theorem | High — your break is here, rebuilt from the picture |
| E6.2 | Right-angled triangles: sin, cos, tan | Low — one decision, then one line of algebra |
| E6.3 | Exact trigonometric values | High — pure Paper 2, and derivable, not memorised |
| E6.4 | Trigonometric functions and graphs | Medium — the second solution is the dropped mark |
| E6.5 | Non-right-angled triangles | Low — once you can pick the right rule |
| E6.6 | Pythagoras and trigonometry in 3D | Low — it is 2D twice, drawn carefully |
Paper 2 is non-calculator: 2 hours, 100 marks, 50% of your grade. Everything in sections 1, 3, 4 and 6 is done by hand and left exact. Sections 2 and 5 genuinely need numerical values of sin, cos and tan, so those are Paper 4 material — there the questions in this guide give you the value you need, because the mark is for the method, not for finding a number in a table.
Number sense and the four operations: solid on all seven rungs. Not one gap, from primary arithmetic right up to Extended. That is the strand everything else stands on, and yours is finished.
Trigonometry rewards that more than almost any topic. Squaring, adding, subtracting, halving, spotting that 25 − 9 is 16 and that 16 is a perfect square — that is most of what a non-calculator Pythagoras question actually asks of you. The arithmetic will not be what loses the mark. Reading the triangle will.
Every idea appears four times, with less help each time.
Where an answer is a surd, type it as 2root13 or 2√13 — both are accepted.
Section 7 is a mixed set: fifteen questions, unlabelled and out of order, because in an exam nobody tells you
whether the question wants Pythagoras, SOHCAHTOA or the cosine rule. Choosing is the skill.
Your check put this strand at step 5 and broke it at step 6. Almost nobody who breaks here has forgotten
a² + b² = c². What goes wrong is upstream of the formula, in the reading of the diagram.
So this section starts there and does not touch the formula for a while.
The hypotenuse is the side opposite the right angle. Not the bottom one, not the longest-looking one, not the one drawn along the page. Opposite the right angle. Find the little square, look straight across from it, and that is your hypotenuse. It happens to also be the longest side, always — but use the right angle to find it, because a diagram can be drawn at any tilt and your eye can be fooled.
In all three the gold side is the hypotenuse, because in all three it is the side facing the small square. In the third one it is the top side and it does not look especially long. That third picture is the one that costs marks.
It is a statement about areas, not about lengths. Draw a square on each of the three sides. The two smaller squares, put together, have exactly the area of the big one.
Written with letters, with c always the hypotenuse:
Everything else follows from rearranging that one line.
There are only two kinds of Pythagoras question, and they use opposite operations. Deciding which you are in takes one glance and saves the whole question.
| What you are given | What you want | What you do | Sanity check |
|---|---|---|---|
| Both short sides | The hypotenuse | Square, add, square-root | Answer must be bigger than both |
| Hypotenuse and one short side | The other short side | Square, subtract, square-root | Answer must be smaller than the hypotenuse |
c on the side opposite the right angle, then a and b on the other two,
in any order. Now the question answers itself: if c is the unknown you add, if c is given
you subtract. You have converted a thinking problem into a looking problem.3, 4, 5 · 5, 12, 13 · 8, 15, 17 · 7, 24, 25 ·
9, 40, 41 · 20, 21, 29. Any multiple works too, so 6-8-10, 9-12-15 and 10-24-26 are
the same three triangles scaled up.Pythagoras links three sides. It cannot touch an angle. The moment an angle appears in the question or in the answer, you need sine, cosine or tangent instead. That is the whole trigger: three sides → Pythagoras; angle involved → SOHCAHTOA.
The hypotenuse is fixed: opposite the right angle, always. The other two names depend on which angle you are standing at, and this is where most of the confusion lives. Pick your angle first, then name the sides from where you are standing.
Same triangle, same three sides, but the labels for opposite and adjacent have swapped, because the angle moved. The hypotenuse did not move, because the right angle did not move.
Choosing between them is mechanical. Mark the side you are given and the side you want; ignore the third side completely. Two out of three letters have now been chosen for you, and only one ratio contains both.
| Given | Want | Ratio |
|---|---|---|
| Hypotenuse | Opposite | sin |
| Hypotenuse | Adjacent | cos |
| Adjacent | Opposite | tan |
| Opposite | Hypotenuse | sin |
| Adjacent | Hypotenuse | cos |
| Opposite | Adjacent | tan |
sin 35° needs a table or a
machine, so questions of that kind belong to Paper 4. The examinable skill on Paper 2 is everything except
that last button: naming the sides, picking the ratio, writing the equation, rearranging it. So in the worked
examples below, where a numerical value of a sine or a tangent is needed, the question simply tells you it.
Your job is the method around it. Where the angle is 30°, 45° or 60°, no value is needed at all —
section 3 gives you those exactly, and those questions are Paper 2.Both are measured from the horizontal, never from a wall or a mast. Elevation is how far you look up; depression is how far you look down.
Draw a north line at every point the journey turns, and mark each bearing clockwise from its own north line. Then look for the right-angled triangle. Work one triangle at a time and keep exact values (surds) until the end.
The shortest distance from a point to a line is the perpendicular distance. Any other path from the point to the line is the hypotenuse of a right-angled triangle that has the perpendicular as one of its sides, so it is longer.
This sub-topic exists for one reason, and it is worth saying plainly: it is how Cambridge examines trigonometry without a calculator. Everything in E6.3 lands on Paper 2, which is half your grade. If a question involves 0°, 30°, 45°, 60° or 90°, no machine is required and none is allowed — the values are exact numbers you are expected to produce.
Most students memorise the table, then lose it under pressure and guess. You are not going to do that. You are going to derive it from two triangles, which takes about twenty seconds and cannot be misremembered, because it is just Pythagoras on shapes you already understand.
Start with an equilateral triangle of side 2. Every angle is 60°. Cut it straight down the middle.
The cut halves the base, so the short side is 1 and the hypotenuse is still 2. Pythagoras gives the third side:
2² − 1² = 4 − 1 = 3, so it is √3. The cut also halves the
60° angle at the top into 30°. Now just read the ratios off, standing at each angle in turn.
Start with a square of side 1 and cut it along a diagonal. The 90° corner splits into two 45° angles,
and the diagonal is √(1 + 1) = √2.
Sine and cosine of 45° are equal because the triangle is symmetric — the opposite and adjacent sides are the same length. That is a fact you can rebuild in a second rather than recall.
These two do not come from a triangle, they come from squashing one. Imagine the angle shrinking towards 0°:
the opposite side vanishes while the hypotenuse and adjacent become the same, so sin 0° = 0
and cos 0° = 1. Open it out towards 90° and the roles swap.
| 0° | 30° | 45° | 60° | 90° | |
|---|---|---|---|---|---|
| sin | 0 | 1⁄2 | √2⁄2 | √3⁄2 | 1 |
| cos | 1 | √3⁄2 | √2⁄2 | 1⁄2 | 0 |
| tan | 0 | √3⁄3 | 1 | √3 | undefined |
0, ½,
√2⁄2, √3⁄2,
1 — the sine row exactly. The cosine row is the same list backwards. Use it as a check on the
triangles, not as a replacement for them.sin 30° must be the small value,
½. A bigger angle has a bigger sine, all the way up to 90°. If your sin 30 came out bigger
than your sin 60, you have them the wrong way round.1⁄√3 and
√3⁄3 are the same number. To go from the first to the second,
multiply top and bottom by √3: (1 × √3) ÷ (√3 × √3) =
√3 ÷ 3. That is called rationalising the denominator, and Cambridge accepts either form —
but you should be able to move between them, because a mark scheme may print one and your working the other.Up to now sine, cosine and tangent have been ratios inside a triangle, so the angle had to be between 0° and 90°. They are also functions, defined for every angle, and drawing them tells you something a triangle cannot: how many angles give the same value.
That matters because of one very specific dropped mark. Ask a machine for the angle whose sine is 0.5 and it returns 30° and stops. Between 0° and 360° there are two answers, and the second one carries a mark of its own. Cambridge asks for it constantly, and it is missed constantly.
−1 ≤ sin x ≤ 1. If your working produces sin x = 1.4, the working is wrong, not the question.x = 90°. The blue line shows why there are two answers: it cuts the curve twice, at 30° and at 150°, and 150 = 180 − 30.The same wave shifted 90° to the left: it starts at 1, drops to 0 at 90°, to −1 at 180°,
back to 0 at 270° and to 1 at 360°. Same range, same period of 360°. Its symmetry is different though
— it is symmetric about x = 180°, so its pair of solutions are x and
360 − x.
Tangent behaves quite differently. It has no maximum and no minimum — it runs off to infinity
— and it breaks at 90° and 270°, where the red dashed lines are. Those are asymptotes: lines
the curve approaches but never touches. It is undefined there because at 90° the adjacent side has shrunk to
zero and you would be dividing by nothing. And its period is 180°, not 360°, so its pair of
solutions are x and x + 180.
| Equation | First solution | Second solution in 0° to 360° | Why |
|---|---|---|---|
| sin x = k | x | 180 − x | the sine curve is symmetric about 90° |
| cos x = k | x | 360 − x | the cosine curve is symmetric about 180° |
| tan x = k | x | x + 180 | the tangent curve repeats every 180° |
y = k across it. Count the crossings. Then read off the second one using the symmetry you can see
in your own sketch. A sketch takes fifteen seconds and turns a remembered rule into a visible fact — which
matters, because remembered rules are exactly what the check showed going wrong under confidence.0° ≤ x ≤ 360°, that range is a direct instruction to look for more than one answer.
Every time you see a range written out like that, sketch the curve.k is one of the exact values — 0, ½,
√2/2, √3/2, 1 — the whole question is Paper 2 material. sin x = √3/2 gives
60° and 120°; cos x = √2/2 gives 45° and 315°. Section 3 plus this table is the
entire method.Rearrange to sin x = k, cos x = k or tan x = k first. If k is negative, find the reference angle from the positive value, then place the answers where the function is negative:
| function | negative for | so the answers are |
|---|---|---|
| sin | 180° to 360° | 180° + ref and 360° − ref |
| cos | 90° to 270° | 180° − ref and 180° + ref |
| tan | 90° to 180° and 270° to 360° | 180° − ref and 360° − ref |
Everything so far has needed a right angle. Take it away and Pythagoras and SOHCAHTOA both stop working. Three formulae replace them, and the only real skill is choosing which one from what you have been given.
Capital letters for angles, lower-case for sides. Side a is opposite angle A,
side b opposite angle B, side c opposite angle C. Every
formula below assumes it.
Note what each one needs. The sine rule works in pairs — a side and the angle facing it, twice over. The cosine rule needs three of the four things in it. The area formula needs two sides and the angle squeezed between them, and that angle must be the included one, not just any angle in the triangle.
| What you are told | What you want | Use |
|---|---|---|
| Two angles and any side | Another side | Sine rule |
| Two sides and an angle opposite one of them | Another angle | Sine rule |
| Two sides and the angle between them | The third side | Cosine rule |
| All three sides | Any angle | Cosine rule |
| Two sides and the angle between them | The area | ½ ab sin C |
−2bc cos A term when the angle is obtuse.
If A is more than 90°, its cosine is negative, and subtracting a negative adds.
So with cos 120° = −½ the term −2bc × (−½)
becomes +bc. Students who subtract anyway get an answer smaller than it should be, and the
error is invisible unless you check. Your diagnostic flagged recurring sign errors across several strands —
this is where they will cost you most, because it is a four-mark question every time.cos 120° = −cos 60° = −½
and cos 150° = −cos 30° = −√3/2. Sine does not flip:
sin 150° = sin 30° = ½. So a triangle containing 30°, 45°, 60°,
120° or 150° is fully workable by hand, and Paper 2 uses exactly those. Anything else — a 37°
angle, say — needs a numerical value of sine or cosine and therefore belongs to Paper 4. Below,
where such a value is needed, the question gives it to you, because the marks are for the method.When all three sides are known, rearrange the cosine rule. The side on top with the minus sign is the side opposite the angle you want:
When the sine rule gives an angle, it gives sin B = k, and two angles between 0° and 180° have that sine: B and 180° − B. Both are possible whenever the second one still leaves room in the triangle, that is when A + (180° − B) is less than 180°. Then two different triangles fit the information. It happens when you are given two sides and an angle that is not between them, opposite the shorter side.
C must be the angle between sides a and b. Backwards, rearrange to sin C = 2 × area ÷ ab; the question tells you whether the angle is acute or obtuse.
Three-dimensional questions look far harder than they are. Nothing new is introduced — there is no 3D version of Pythagoras and no 3D trigonometry. There is only one technique, and it is this:
Once it is redrawn, the question is an ordinary 2D one from section 1 or section 2. Almost every lost mark in E6.6 comes from trying to work inside the 3D picture instead of pulling a flat triangle out of it.
A cuboid has two different diagonals and they are easy to confuse. The face diagonal lies flat on one face. The space diagonal cuts through the inside from one corner to the opposite corner. You get the second by using the first, in two steps.
Step 1 uses the rectangle on the floor: the base diagonal is the hypotenuse of a triangle with the length and the width as its short sides. Step 2 uses a vertical triangle standing on that diagonal, with the height going straight up. Here is step 2 pulled out and drawn flat, which is what you should physically do on the page:
d² = (l² + w²) + h². So the space diagonal of a cuboid is
√(l² + w² + h²). Examiners still want to see the two-step working, but the
one-liner is an excellent check on the answer.Nothing here is labelled. That is deliberate: in an exam nobody tells you whether a question wants Pythagoras, SOHCAHTOA, an exact value, a second solution or the cosine rule. Deciding is the skill that blocked practice never builds, because in blocked practice you already know the answer to that question.
Before each one, ask yourself two things. Is there a right angle? And is an angle involved, or only sides? Those two answers pick the method almost every time.
Seven things. Between them they cover nearly everything an examiner can ask in topic 6.
c there before you write a single number.180 − x, cosine takes 360 − x, tangent takes x + 180.Come back to section 7 in a week without reading anything above it. A high score on a cold run means the topic is set. Anything you miss will point at exactly one section to reopen — and it will be one section, not the whole topic.