Solutions — Surds & Trig Unit Summary Worksheet
Year 10 Core Mathematics · Cambridge Ch 4 (Surds) & Ch 6 (Trigonometry) · Mr Wong
ANSWER KEY
Section 1 — Simplifying surds
Q1. Pull out the biggest square factor:
a $4\sqrt{3}$ ($\sqrt{16\cdot 3}$)
b $5\sqrt{3}$ ($\sqrt{25\cdot 3}$)
c $10\sqrt{2}$ ($\sqrt{100\cdot 2}$)
d $7\sqrt{2}$ ($\sqrt{49\cdot 2}$)
Section 2 — Adding & subtracting
Q2. Simplify, then combine like surds:
a $7\sqrt{3}$
b $5\sqrt{3}$ ($2\sqrt{3}+3\sqrt{3}$)
c $3\sqrt{2}$ ($5\sqrt{2}-2\sqrt{2}$)
d $2\sqrt{5}$ ($4\sqrt{5}-2\sqrt{5}$)
Section 3 — Multiplying, dividing & brackets
Q3. Combine the insides, then simplify:
a $3\sqrt{2}$ ($\sqrt{18}$)
b $2\sqrt{2}$ ($\sqrt{8}$)
c $5$
d $1$ ($4-3$)
Q4. Expand $(3+\sqrt{2})^{2}$:
· $11+6\sqrt{2}$ ($9+3\sqrt{2}+3\sqrt{2}+2$)
Section 4 — Rationalising
Q5. Multiply top & bottom by the denominator surd:
a $\sqrt{3}$ ($\tfrac{3\sqrt{3}}{3}$)
b $5\sqrt{2}$ ($\tfrac{10\sqrt{2}}{2}$)
c $\sqrt{3}$ ($\tfrac{6\sqrt{3}}{6}$)
d $2\sqrt{2}$ ($\tfrac{4\sqrt{2}}{2}$)
Section 5 — Finding a side
Q6. SOHCAHTOA (to 2 d.p.):
a $6.43$ ($10\sin 40°$)
b $13.24$ ($15\cos 28°$)
c $12.85$ ($9\tan 55°$)
Section 6 — Finding an angle
Q7. Inverse trig (to 1 d.p.):
a $35.7°$ ($\sin^{-1}\tfrac{7}{12}$)
b $56.9°$ ($\cos^{-1}\tfrac{6}{11}$)
c $24.0°$ ($\tan^{-1}\tfrac{4}{9}$)
Section 7 — Elevation & depression
Q8. Tree height $=$ eye height $+$ rise above eye level:
·
$1.6+20\tan 35°=1.6+14.00=15.60$. Height $\approx 15.6$ m
Section 8 — Bearings
Q9. $040°$ is $40°$ clockwise of north; the $15$ km is the hypotenuse:
a north $\approx 11.49$ km ($15\cos 40°$)
b east $\approx 9.64$ km ($15\sin 40°$)
Challenge — 3D box
Q10. Base $6\times 8$, height $5$:
a
Space diagonal $=\sqrt{6^{2}+8^{2}+5^{2}}=\sqrt{125}=5\sqrt{5}$. $5\sqrt{5}\approx 11.18$ cm
b
Base diagonal $=\sqrt{36+64}=10$, so angle $=\tan^{-1}\!\big(\tfrac{5}{10}\big)$. $\approx 26.6°$