Year 10 Core — Surds & Trigonometry: Challenge questions

Cambridge Core, Chapter 4 (Surds) & Chapter 6 (Trigonometry)  •  extension · harder multi-step problems  •  Mr Wong
📚 Linked pages: 📄 Challenge worksheet ✅ Solutions (answer key) ← Back to the unit summary

How to use this page

These are harder than the standard worksheet — the kind of multi-step problems that separate an A from a B, and a good bridge into Year 11 Methods. Every step you'd actually write is shown in the solution. The rule for using this page:

Try the whole question first — pencil and paper, no peeking. Stuck after a genuine attempt? Hit Hint. Only open Show full solution once you've committed to an answer, then mark your own working line by line. The stars show difficulty: ★★ solid extension, ★★★ properly challenging.

Part A — Surds (harder)

★★ A1 — rationalise a binomial denominator
Rationalise the denominator and simplify fully: $\dfrac{3}{\sqrt{5}-\sqrt{2}}$.
When the bottom is a difference like $\sqrt5-\sqrt2$, multiply top and bottom by its conjugate $\sqrt5+\sqrt2$. The bottom becomes $(\sqrt5)^2-(\sqrt2)^2$ — the surds disappear.
  1. Multiply top and bottom by the conjugate $\sqrt5+\sqrt2$ (this is multiplying by $1$): $\dfrac{3}{\sqrt5-\sqrt2}\times\dfrac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}$.
  2. The bottom is a difference of squares: $(\sqrt5-\sqrt2)(\sqrt5+\sqrt2)=5-2=3$.
  3. So we have $\dfrac{3(\sqrt5+\sqrt2)}{3}$.
  4. Cancel the $3$: $\sqrt5+\sqrt2$.

$\dfrac{3}{\sqrt5-\sqrt2}=$ $\sqrt5+\sqrt2$

★★ A2 — express in the form $a+b\sqrt3$
Express $\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$ in the form $a+b\sqrt{3}$, where $a$ and $b$ are integers.
Multiply top and bottom by the conjugate of the denominator, $\sqrt3+1$. The numerator becomes a perfect square: $(\sqrt3+1)^2$.
  1. Multiply top and bottom by $\sqrt3+1$: $\dfrac{(\sqrt3+1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$.
  2. Bottom: $(\sqrt3)^2-1^2=3-1=2$.
  3. Top: $(\sqrt3+1)^2=3+2\sqrt3+1=4+2\sqrt3$.
  4. So $\dfrac{4+2\sqrt3}{2}=\dfrac{4}{2}+\dfrac{2\sqrt3}{2}=2+\sqrt3$.

$\dfrac{\sqrt3+1}{\sqrt3-1}=$ $2+\sqrt3$  ($a=2,\ b=1$)

★★ A3 — combine three terms
Simplify fully: $\sqrt{75}-\sqrt{12}+\dfrac{6}{\sqrt{3}}$.
Three different-looking terms, but every one is secretly a multiple of $\sqrt3$. Simplify the first two, rationalise the third, then add.
  1. $\sqrt{75}=\sqrt{25\times3}=5\sqrt3$.
  2. $\sqrt{12}=\sqrt{4\times3}=2\sqrt3$.
  3. $\dfrac{6}{\sqrt3}=\dfrac{6}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{6\sqrt3}{3}=2\sqrt3$.
  4. Now they are all like surds: $5\sqrt3-2\sqrt3+2\sqrt3=5\sqrt3$.

$\sqrt{75}-\sqrt{12}+\dfrac{6}{\sqrt3}=$ $5\sqrt3$

★★ A4 — expand a square
Expand and simplify $\big(2\sqrt{5}-3\big)^{2}$.
Use $(p-q)^2=p^2-2pq+q^2$ with $p=2\sqrt5$ and $q=3$. Remember $(2\sqrt5)^2=4\times5$.
  1. $(2\sqrt5)^2=2^2\times(\sqrt5)^2=4\times5=20$.
  2. Middle term: $-2\times(2\sqrt5)\times3=-12\sqrt5$.
  3. Last term: $(-3)^2=9$.
  4. Collect the whole numbers: $20+9=29$.

$\big(2\sqrt5-3\big)^2=$ $29-12\sqrt5$

★★ A5 — surds inside Pythagoras
A right-angled triangle has hypotenuse $\sqrt{50}$ cm and one shorter side $\sqrt{18}$ cm. Find the exact length of the third side, in simplest surd form.
Pythagoras: $\text{(third side)}^2=\text{hyp}^2-\text{known}^2$. You don't even need to simplify the surds first — squaring a square root just removes it.
  1. Let the third side be $x$. By Pythagoras, $x^2+\big(\sqrt{18}\big)^2=\big(\sqrt{50}\big)^2$.
  2. Squaring removes the roots: $x^2+18=50$.
  3. $x^2=32$, so $x=\sqrt{32}$.
  4. Simplify: $\sqrt{32}=\sqrt{16\times2}=4\sqrt2$.

Third side $=$ $4\sqrt2$ cm

Nice structure: $\sqrt{50}=5\sqrt2$, $\sqrt{18}=3\sqrt2$, third side $4\sqrt2$ — a $3$-$4$-$5$ triangle scaled by $\sqrt2$.

★★★ A6 — show, then "hence"
(a) Show that $\big(\sqrt{7}+\sqrt{3}\big)\big(\sqrt{7}-\sqrt{3}\big)=4$.   (b) Hence simplify $\dfrac{1}{\sqrt{7}+\sqrt{3}}+\dfrac{1}{\sqrt{7}-\sqrt{3}}$.
For (b), put the two fractions over the common denominator from part (a). The "hence" is a strong signal: the answer to (a) is the denominator you need.
  1. (a) Difference of squares: $\big(\sqrt7+\sqrt3\big)\big(\sqrt7-\sqrt3\big)=(\sqrt7)^2-(\sqrt3)^2=7-3=4$. ✓
  2. (b) Common denominator is $\big(\sqrt7+\sqrt3\big)\big(\sqrt7-\sqrt3\big)=4$ from part (a).
  3. $\dfrac{1}{\sqrt7+\sqrt3}+\dfrac{1}{\sqrt7-\sqrt3}=\dfrac{\big(\sqrt7-\sqrt3\big)+\big(\sqrt7+\sqrt3\big)}{4}$.
  4. The $\sqrt3$ terms cancel on top: $\dfrac{2\sqrt7}{4}=\dfrac{\sqrt7}{2}$.

$\dfrac{1}{\sqrt7+\sqrt3}+\dfrac{1}{\sqrt7-\sqrt3}=$ $\dfrac{\sqrt7}{2}$

★★★ A7 — solve a surd equation
Solve $\sqrt{x}=\sqrt{x-5}+1$ for $x$.
Don't expand blindly. Square both sides once — the $\sqrt{x-5}$ survives, but the lone $\sqrt x$ disappears. Rearrange so the remaining root is alone, then square again.
  1. Square both sides: $x=\big(\sqrt{x-5}+1\big)^2=(x-5)+2\sqrt{x-5}+1$.
  2. Simplify the right: $x=x-4+2\sqrt{x-5}$.
  3. Subtract $x$ from both sides: $0=-4+2\sqrt{x-5}$, so $2\sqrt{x-5}=4$.
  4. Divide by $2$: $\sqrt{x-5}=2$. Square again: $x-5=4$, so $x=9$.
  5. Check (always check after squaring): $\sqrt9=3$ and $\sqrt{9-5}+1=\sqrt4+1=2+1=3$. ✓

$x=9$

⚠️ Squaring can invent fake solutions, so the final check isn't optional — it's part of the method.

Part B — Trigonometry (harder)

★★★ B1 — two observers, one tower
From a point $A$ on level ground the angle of elevation to the top of a vertical tower is $30°$. From a point $B$, which is $40$ m closer to the tower along the same straight line, the angle of elevation is $50°$. Find the height of the tower, to 2 decimal places.
A B foot top 30° 50° ←——— 40 m ———→ B then closer
Call the height $h$ and the distance from $B$ to the foot $d$. Write $\tan50=\dfrac{h}{d}$ and $\tan30=\dfrac{h}{d+40}$. Make $d$ the subject of each and set them equal — that gives one equation in $h$.
  1. Let $h$ = tower height, $d$ = distance from $B$ to the foot. Then $A$ is $d+40$ from the foot.
  2. From $B$: $\tan50°=\dfrac{h}{d}\Rightarrow d=\dfrac{h}{\tan50°}$.
  3. From $A$: $\tan30°=\dfrac{h}{d+40}\Rightarrow d+40=\dfrac{h}{\tan30°}$.
  4. Subtract: $\dfrac{h}{\tan30°}-\dfrac{h}{\tan50°}=40$.
  5. Factor out $h$: $h\left(\dfrac{1}{\tan30°}-\dfrac{1}{\tan50°}\right)=40$, i.e. $h\,(1.73205-0.83910)=40$.
  6. $h\times0.89295=40\Rightarrow h=\dfrac{40}{0.89295}=44.795\ldots$

Height $\approx$ $44.80$ m

★★★ B2 — a two-leg journey
A ship sails $12$ km from port $P$ on a bearing of $050°$T, then changes course and sails $16$ km on a bearing of $140°$T to reach $Q$. (a) Show that the ship's two legs are at right angles. (b) Find the straight-line distance $PQ$. (c) Find the bearing of $Q$ from $P$, to the nearest degree.
(a) Compare the two bearings. (b) With a right angle between the legs, $PQ$ is just the hypotenuse — Pythagoras. (c) Find the angle between the first leg and $PQ$ with $\tan$, then add it to $050°$.
  1. (a) The change of direction is $140°-50°=90°$, so the second leg is perpendicular to the first. ✓
  2. (b) The two legs form the right angle of a triangle, so $PQ$ is the hypotenuse: $PQ=\sqrt{12^2+16^2}=\sqrt{144+256}=\sqrt{400}=20$ km.
  3. (c) Let $\theta$ be the angle at $P$ between the first leg and $PQ$. The side opposite $\theta$ is the second leg ($16$), the adjacent is the first leg ($12$): $\tan\theta=\dfrac{16}{12}\Rightarrow\theta=\tan^{-1}\!\big(\tfrac{4}{3}\big)\approx53.13°$.
  4. The second leg turns clockwise (bearing increased $50°\to140°$), so $Q$ is $53.13°$ clockwise of the first leg: bearing $=50°+53.13°=103.13°$.

(b) $PQ=$ $20$ km;   (c) bearing of $Q$ from $P\approx$ $103°$T

★★★ B3 — square-based pyramid (two triangles)
A right pyramid has a square base of side $10$ cm and its apex is directly above the centre of the base, $12$ cm up. Find (a) the length of a slant edge (apex to a base corner), exact and to 2 d.p.; (b) the angle a slant edge makes with the base, to 1 d.p.
First find the horizontal distance from the centre of the base to a corner — that's half the base diagonal. Then the height, that half-diagonal, and the slant edge form a right-angled triangle.
  1. Base diagonal $=\sqrt{10^2+10^2}=\sqrt{200}=10\sqrt2$, so half of it (centre to corner) is $5\sqrt2\approx7.071$ cm.
  2. (a) The slant edge is the hypotenuse of the right triangle with legs $12$ (height) and $5\sqrt2$: slant $=\sqrt{12^2+(5\sqrt2)^2}=\sqrt{144+50}=\sqrt{194}\approx13.93$ cm.
  3. (b) The angle $\alpha$ with the base uses height (opposite) over half-diagonal (adjacent): $\tan\alpha=\dfrac{12}{5\sqrt2}=\dfrac{12}{7.071}\Rightarrow\alpha=\tan^{-1}(1.6971)\approx59.5°$.

(a) slant edge $=$ $\sqrt{194}\approx13.93$ cm;   (b) angle $\approx$ $59.5°$

★★★ B4 — drone with two angles of depression
A drone hovers $80$ m directly above a straight road. Looking down the road, the angle of depression to a parked car is $25°$, and to a person standing further along (beyond the car) is $15°$. Find the distance between the car and the person, to 2 d.p.
Angle of depression to a point = the angle of elevation from that point back up. Each gives a right triangle of height $80$. Find each horizontal distance, then subtract.
  1. For the car: $\tan25°=\dfrac{80}{\text{(distance to car)}}\Rightarrow \text{car}=\dfrac{80}{\tan25°}=171.56$ m.
  2. For the person (smaller angle ⇒ further away): $\text{person}=\dfrac{80}{\tan15°}=298.56$ m.
  3. Both are measured from the point on the road directly below the drone, so the gap between them is the difference: $298.56-171.56=127.00$ m.

Distance car → person $\approx$ $127.00$ m

A smaller angle of depression always means the object is further away — a good sanity check on which distance should be larger.

★★ B5 — surds meet trig
An equilateral triangle has side length $2\sqrt{3}$ cm. Find its exact area (leave your answer in surd form).
Drop a perpendicular from one vertex to split the triangle into two right-angled triangles. The height comes out as a surd; then area $=\tfrac12\times\text{base}\times\text{height}$. (Or use area $=\tfrac{\sqrt3}{4}s^2$.)
  1. Split the triangle down the middle. Each right triangle has base $\dfrac{2\sqrt3}{2}=\sqrt3$ and hypotenuse $2\sqrt3$.
  2. Height $h$ by Pythagoras: $h=\sqrt{(2\sqrt3)^2-(\sqrt3)^2}=\sqrt{12-3}=\sqrt9=3$.
  3. Area $=\tfrac12\times\text{base}\times\text{height}=\tfrac12\times2\sqrt3\times3=3\sqrt3$.
  4. (Check with the formula $\tfrac{\sqrt3}{4}s^2=\tfrac{\sqrt3}{4}\times12=3\sqrt3$. ✓)

Area $=$ $3\sqrt3$ cm²

One more for the road

Stretch: In question A6 you found $\dfrac{1}{\sqrt7+\sqrt3}+\dfrac{1}{\sqrt7-\sqrt3}=\dfrac{\sqrt7}{2}$. Can you predict, without a calculator, what $\dfrac{1}{\sqrt{11}+\sqrt{7}}+\dfrac{1}{\sqrt{11}-\sqrt{7}}$ will simplify to? (Try it, then check: the same method gives $\dfrac{2\sqrt{11}}{11-7}=\dfrac{2\sqrt{11}}{4}=\dfrac{\sqrt{11}}{2}$.)

Worked the lot? You're ready for the test — and you've already met ideas that come back in Year 11 Methods.  — Mr Wong