Year 10 Core — Surds & Trigonometry (Unit summary)

Cambridge Core, Chapter 4 (Surds) & Chapter 6 (Trigonometry)  •  revision · cheat sheet · mixed practice  •  Mr Wong
📚 Also for this revision: 📄 Mixed-practice worksheet ✅ Solutions (answer key) ↓ Deep dive into a sub-topic

What this page is

One-stop revision for the whole Surds and Trigonometry unit. Use it the night before a test:

Part 1 — Cheat sheet

§4A–4B · Simplifying & like surds

Make the number under the root small

A surd is a root that doesn't come out to a whole number (e.g. $\sqrt{2}$ is a surd, $\sqrt{9}=3$ is not).

Pull out the biggest square factor: $\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$.

Add / subtract only like surds: $3\sqrt{2}+5\sqrt{2}=8\sqrt{2}$. Simplify first so you can see the like ones.

$\sqrt{a}+\sqrt{b}\ne\sqrt{a+b}$. You can't add the insides.

§4C · Multiplying & dividing

Combine the insides

$\sqrt{x}\times\sqrt{y}=\sqrt{xy}$,   $\dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}$

$\sqrt{x}\times\sqrt{x}=x$   (a surd times itself loses the root).

Expand brackets with the distributive law, just like with letters: $(2+\sqrt{3})(2-\sqrt{3})=4-3=1$.

Always simplify the answer: $\sqrt{6}\times\sqrt{3}=\sqrt{18}=3\sqrt{2}$.

§4D · Rationalising

No surd in the denominator

Multiply top and bottom by the surd on the bottom: $\dfrac{6}{\sqrt{3}}=\dfrac{6}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{6\sqrt{3}}{3}=2\sqrt{3}$.

For $\dfrac{a}{b\sqrt{y}}$, multiply by $\dfrac{\sqrt{y}}{\sqrt{y}}$ — the bottom becomes $b\,y$ (a whole number).

Multiplying by $\tfrac{\sqrt{y}}{\sqrt{y}}$ is multiplying by $1$, so the value never changes — only the look.

§6A · Finding a side

SOHCAHTOA

$\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$,   $\cos\theta=\dfrac{\text{adj}}{\text{hyp}}$,   $\tan\theta=\dfrac{\text{opp}}{\text{adj}}$

Label hyp (opposite the right angle), then opp and adj relative to your angle.

Pick the ratio that uses the two sides you care about, then solve.

Unknown on top → multiply. Unknown on the bottom → divide. Keep your calculator in degrees.

§6B · Finding an angle

Inverse trig

Know two sides, want the angle → use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$ (the SHIFT / 2nd key).

e.g. $\tan\theta=\dfrac{5}{8}\Rightarrow\theta=\tan^{-1}\!\big(\tfrac{5}{8}\big)\approx 32.0°$.

Choose the ratio from the two sides you were given, then take the inverse of that ratio.

§6C–6E · Applications

Draw the triangle first

Elevation = angle up from horizontal; depression = angle down from horizontal.

Bearings: clockwise from north, 3 digits, e.g. $060°$T. The reverse bearing differs by $180°$.

3D: find the right-angled triangle inside the solid, redraw it flat, use Pythagoras first if you need a missing side.

Always answer in a sentence with units, and round only at the end.

Part 2 — Which technique?

🧭 Decision tree — pick a tool by question type

  1. Number under a root, like $\sqrt{72}$?  → pull out the biggest square factor to simplify. (§4A)
  2. Adding or subtracting surds?  → simplify each one first, then combine only the like surds. (§4B)
  3. Multiplying or dividing surds?  → $\sqrt{x}\times\sqrt{y}=\sqrt{xy}$, then simplify the result. (§4C)
  4. Brackets with surds inside?  → expand with the distributive law, remember $\sqrt{x}\times\sqrt{x}=x$. (§4C)
  5. A surd on the bottom of a fraction?  → rationalise: multiply top and bottom by that surd. (§4D)
  6. Right triangle, know an angle and a side, want another side?  → SOHCAHTOA — pick the ratio with your two sides. (§6A)
  7. Right triangle, know two sides, want the angle?  → inverse trig ($\sin^{-1}/\cos^{-1}/\tan^{-1}$). (§6B)
  8. Word problem with "angle of elevation / depression"?  → draw and label a right-angled triangle, then choose a ratio. (§6C)
  9. "Bearing of B from A" or a navigation distance?  → measure clockwise from north at A; split the trip into north/south and east/west legs. (§6D)
  10. A 3D solid (box, pyramid, mast)?  → find the hidden right triangle, redraw it flat, Pythagoras first if needed, then trig. (§6E)

Part 3 — Mixed worked examples

Each example crosses one or two sub-topics. Try it first, then click Show solution.

§4A–4B EXAMPLE 1 — simplify, then subtract
Simplify $\sqrt{72}-\sqrt{18}$.
  1. Simplify each surd with its biggest square factor: $\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$,   $\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}$.
  2. Now they are like surds, so subtract the numbers in front: $6\sqrt{2}-3\sqrt{2}=3\sqrt{2}$.

$\sqrt{72}-\sqrt{18}=\boxed{3\sqrt{2}}$.

Why simplify first? Before simplifying they look different ($\sqrt{72}$ vs $\sqrt{18}$); after, you can see they're both "$\sqrt{2}$" surds and combine.

§4C EXAMPLE 2 — expand brackets
Expand and simplify $(3+\sqrt{2})^{2}$.

Write it out and use the distributive law (or "FOIL"): $(3+\sqrt{2})(3+\sqrt{2})$.

$=3\times 3+3\sqrt{2}+3\sqrt{2}+\sqrt{2}\times\sqrt{2}=9+6\sqrt{2}+2$.

Remember $\sqrt{2}\times\sqrt{2}=2$. Collect the whole numbers: $9+2=11$.

$(3+\sqrt{2})^{2}=\boxed{11+6\sqrt{2}}$.

§4D EXAMPLE 3 — rationalise the denominator
Rationalise $\dfrac{10}{\sqrt{5}}$.

Multiply top and bottom by $\sqrt{5}$ (that's multiplying by $1$, so the value is unchanged):

$\dfrac{10}{\sqrt{5}}\times\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{10\sqrt{5}}{5}$.

Now simplify the whole-number fraction $\tfrac{10}{5}=2$:

$\dfrac{10}{\sqrt{5}}=\boxed{2\sqrt{5}}$.

§6A EXAMPLE 4 — find an unknown side
In a right-angled triangle the hypotenuse is $12$ cm and one angle is $35°$. Find the side opposite that angle, to 2 d.p.
  1. We have the hypotenuse and want the opposite — that's the pair in SOH: $\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$.
  2. Substitute: $\sin 35°=\dfrac{\text{opp}}{12}$.
  3. The unknown is on top, so multiply: $\text{opp}=12\sin 35°$.
  4. On the calculator (in degrees): $12\sin 35°=6.8829\ldots$

$\text{opp}\approx\boxed{6.88\text{ cm}}$.

§6B EXAMPLE 5 — find an unknown angle
A right-angled triangle has the side opposite angle $\theta$ equal to $5$ and the side adjacent to $\theta$ equal to $8$. Find $\theta$ to 1 d.p.
  1. We have opposite and adjacent — that's the pair in TOA: $\tan\theta=\dfrac{\text{opp}}{\text{adj}}=\dfrac{5}{8}$.
  2. To get the angle, take the inverse: $\theta=\tan^{-1}\!\big(\tfrac{5}{8}\big)$.
  3. On the calculator: $\theta=32.0054\ldots°$

$\theta\approx\boxed{32.0°}$.

⚠️ Use the inverse key ($\tan^{-1}$, usually SHIFT then tan) — not $1\div\tan$. They are different things.

§6D EXAMPLE 6 — bearings (two-step)
A hiker walks $8$ km from camp on a bearing of $060°$T. How far east and how far north of camp are they? (2 d.p.)
  1. Draw north pointing up. A bearing of $060°$ is $60°$ clockwise from north, so the angle between the path and the north line is $60°$.
  2. The $8$ km is the hypotenuse. The north leg is alongside the north line (adjacent to the $60°$): $\text{north}=8\cos 60°=4.00$ km.
  3. The east leg is across from the $60°$ (opposite): $\text{east}=8\sin 60°=6.9282\ldots\approx 6.93$ km.

They are $\boxed{6.93\text{ km east}}$ and $\boxed{4.00\text{ km north}}$ of camp.

Tip: the angle you measure from the north line decides which leg gets $\cos$ (along north) and which gets $\sin$ (across to east). Draw it every time.

Part 4 — Deep dive into a sub-topic

Pick a section and jump to its full lesson, worksheet and solutions.

§4A–4BSimplifying & like surdsRational vs irrational, square factors, add/subtract §4CMultiplying & dividingCombine insides, squares of surds, brackets §4DRationalisingClear a surd from the denominator §6AFinding a sideSOHCAHTOA, labelling, choosing a ratio §6BFinding an angleInverse trig from two known sides §6CApplications in 2DAngles of elevation and depression §6DDirections & bearingsTrue bearings, reverse bearings, N/S/E/W legs §6EApplications in 3DFind the right triangle inside a solid

Part 5 — Watch this if you're still hazy

Four short walkthroughs that anchor the most-used ideas in the unit — two from surds, two from trig.

📺 Simplifying a surd: pulling the biggest square factor out of $\sqrt{50}$ to get $5\sqrt{2}$ — the move behind almost every Chapter 4 question.

📺 Rationalising the denominator: why multiplying top and bottom by the bottom surd clears the root without changing the value.

📺 SOHCAHTOA in action: labelling the sides, choosing the right ratio, and solving for an unknown side.

📺 Bearings into a triangle: turning a navigation bearing into a right-angled triangle to find how far north and east you've travelled.

Part 6 — Unit quiz (10 Q · 5 min · auto-marked)

Pick the correct answer for each, then click Mark.

Q1. Which of these is a surd (an irrational number)?

Q2. Simplify $\sqrt{50}$.

Q3. $3\sqrt{2}+5\sqrt{2}=$

Q4. $\sqrt{3}\times\sqrt{12}=$

Q5. $(2+\sqrt{3})(2-\sqrt{3})=$

Q6. Rationalised, $\dfrac{1}{\sqrt{2}}=$

Q7. Which ratio uses the opposite and the hypotenuse?

Q8. To find a side using $\cos 40°=\dfrac{\text{adj}}{\text{hyp}}$ where the hypotenuse is $10$, the adjacent side is:

Q9. If $\tan\theta=1$, then $\theta=$

Q10. Point $B$ is due east of point $A$. The bearing of $B$ from $A$ is:

Working program — full unit

After the quiz, open Cambridge Core Chapters 4 and 6 and run through the following mixed-review sets:

Anything you got wrong on the quiz → click the matching deep-dive link above and revisit that lesson + its worksheet.

Self-check — write in your book

Before you call this revision done, write one line each:
  1. How do you decide which square factor to pull out of a surd?
  2. What do you multiply by to rationalise a denominator, and why doesn't it change the value?
  3. Given two sides and wanting the angle, what do you do after writing the ratio?
  4. For a bearing, how do you decide which leg uses $\cos$ and which uses $\sin$?

— Mr Wong