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 $=\tan^{-1}\!\big(\tfrac{5}{10}\big)$. $\approx 26.6°$