Solutions — Surds & Trig Challenge

Year 10 Core Mathematics · Extension · Cambridge Ch 4 & Ch 6 · Mr Wong

ANSWER KEY

Part A — Surds

A1. Multiply by the conjugate $\sqrt5+\sqrt2$.

$\dfrac{3}{\sqrt5-\sqrt2}\times\dfrac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}=\dfrac{3(\sqrt5+\sqrt2)}{5-2}=\dfrac{3(\sqrt5+\sqrt2)}{3}$

$=\sqrt5+\sqrt2$

A2. Multiply by the conjugate $\sqrt3+1$.

Bottom: $(\sqrt3-1)(\sqrt3+1)=3-1=2$. Top: $(\sqrt3+1)^2=3+2\sqrt3+1=4+2\sqrt3$.

$\dfrac{4+2\sqrt3}{2}=$ $2+\sqrt3$ ($a=2,\ b=1$)

A3. Every term is a multiple of $\sqrt3$.

$\sqrt{75}=5\sqrt3$, $\sqrt{12}=2\sqrt3$, $\dfrac{6}{\sqrt3}=\dfrac{6\sqrt3}{3}=2\sqrt3$.

$5\sqrt3-2\sqrt3+2\sqrt3=$ $5\sqrt3$

A4. $(p-q)^2=p^2-2pq+q^2$ with $p=2\sqrt5,\ q=3$.

$(2\sqrt5)^2-2(2\sqrt5)(3)+3^2=20-12\sqrt5+9=$ $29-12\sqrt5$

A5. Pythagoras; squaring removes the roots.

$x^2+\big(\sqrt{18}\big)^2=\big(\sqrt{50}\big)^2\Rightarrow x^2+18=50\Rightarrow x^2=32$.

$x=\sqrt{32}=$ $4\sqrt2$ cm (a $3$-$4$-$5$ triangle scaled by $\sqrt2$).

A6. (a) $(\sqrt7+\sqrt3)(\sqrt7-\sqrt3)=7-3=4$. ✓

(b) Common denominator $=4$ (from part a):

$\dfrac{(\sqrt7-\sqrt3)+(\sqrt7+\sqrt3)}{4}=\dfrac{2\sqrt7}{4}=$ $\dfrac{\sqrt7}{2}$

A7. Square, isolate the remaining root, square again, then check.

$x=(x-5)+2\sqrt{x-5}+1\Rightarrow x=x-4+2\sqrt{x-5}$.

$4=2\sqrt{x-5}\Rightarrow\sqrt{x-5}=2\Rightarrow x-5=4\Rightarrow$ $x=9$.

Check: $\sqrt9=3$ and $\sqrt4+1=3$. ✓ (No fake solution introduced.)

Part B — Trigonometry · Mr Wong

ANSWER KEY

B1. Let height $=h$, distance from $B$ to the foot $=d$.

From $B$: $d=\dfrac{h}{\tan50°}$. From $A$: $d+40=\dfrac{h}{\tan30°}$.

Subtract: $\dfrac{h}{\tan30°}-\dfrac{h}{\tan50°}=40\Rightarrow h(1.73205-0.83910)=40$.

$h=\dfrac{40}{0.89295}=44.795\ldots\Rightarrow$ height $\approx 44.80$ m.

B2. (a) $140°-50°=90°$, so the legs are perpendicular. ✓

(b) $PQ=\sqrt{12^2+16^2}=\sqrt{400}=$ $20$ km.

(c) Angle at $P$: $\tan\theta=\dfrac{16}{12}\Rightarrow\theta=\tan^{-1}\!\big(\tfrac43\big)=53.13°$.

Turn is clockwise, so bearing $=50°+53.13°=103.13°\approx$ $103°$T.

B3. Centre-to-corner $=$ half the base diagonal $=\dfrac{\sqrt{10^2+10^2}}{2}=\dfrac{10\sqrt2}{2}=5\sqrt2\approx7.071$ cm.

(a) slant $=\sqrt{12^2+(5\sqrt2)^2}=\sqrt{144+50}=\sqrt{194}\approx$ $13.93$ cm.

(b) $\tan\alpha=\dfrac{12}{5\sqrt2}=1.6971\Rightarrow\alpha=\tan^{-1}(1.6971)\approx$ $59.5°$.

B4. Each angle of depression equals the elevation from the ground point; both triangles have height $80$.

Car: $\dfrac{80}{\tan25°}=171.56$ m. Person: $\dfrac{80}{\tan15°}=298.56$ m.

Gap $=298.56-171.56=$ $127.00$ m (smaller angle ⇒ further away ✓).

B5. Split into two right triangles; base $=\sqrt3$, hypotenuse $=2\sqrt3$.

Height $=\sqrt{(2\sqrt3)^2-(\sqrt3)^2}=\sqrt{12-3}=\sqrt9=3$.

Area $=\tfrac12\times2\sqrt3\times3=$ $3\sqrt3$ cm² (check: $\tfrac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3$ ✓).