Surds & Trig Unit Summary — mixed practice

Year 10 Core Mathematics · Cambridge Ch 4 (Surds) & Ch 6 (Trigonometry) · Mr Wong

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Class

Date

Page 1 — Surds (Chapter 4).

§4A–4B simplifying & like surds · §4C multiplying, dividing & brackets · §4D rationalising
Give exact surd answers (no decimals). Always simplify as far as you can.

Quick reference

$\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$ · $\sqrt{x}\times\sqrt{y}=\sqrt{xy}$ · $\sqrt{x}\times\sqrt{x}=x$

$\dfrac{a}{\sqrt{y}}=\dfrac{a}{\sqrt{y}}\times\dfrac{\sqrt{y}}{\sqrt{y}}=\dfrac{a\sqrt{y}}{y}$ (rationalise) · only like surds add

Section 1 — Simplifying surds (§4A)

1. Write each in simplest surd form.

a $\sqrt{48}=$

b $\sqrt{75}=$

c $\sqrt{200}=$

d $\sqrt{98}=$

Section 2 — Adding & subtracting (§4B)

2. Simplify first if needed, then combine the like surds.

a $5\sqrt{3}+2\sqrt{3}=$

b $\sqrt{12}+\sqrt{27}=$

c $\sqrt{50}-\sqrt{8}=$

d $4\sqrt{5}-\sqrt{20}=$

Section 3 — Multiplying, dividing & brackets (§4C)

3. Simplify each.

a $\sqrt{6}\times\sqrt{3}=$

b $\dfrac{\sqrt{40}}{\sqrt{5}}=$

c $\big(\sqrt{5}\big)^{2}=$

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

4. Expand and simplify $(3+\sqrt{2})^{2}$. Answer:

Section 4 — Rationalising the denominator (§4D)

5. Rationalise and simplify.

a $\dfrac{3}{\sqrt{3}}=$

b $\dfrac{10}{\sqrt{2}}=$

c $\dfrac{6}{2\sqrt{3}}=$

d $\dfrac{4}{\sqrt{2}}=$

Surds & Trig Unit Summary — continued

Page 2 — Trigonometry (Chapter 6): sides · angles · elevation/depression · bearings · 3D

Name

Quick reference

$\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$ · $\cos\theta=\dfrac{\text{adj}}{\text{hyp}}$ · $\tan\theta=\dfrac{\text{opp}}{\text{adj}}$ (SOHCAHTOA)

angle from two sides → $\sin^{-1}/\cos^{-1}/\tan^{-1}$ · bearing = clockwise from north, 3 digits · keep calculator in degrees

Section 5 — Finding a side (§6A)

6. Find the marked length to 2 d.p. (calculator in degrees).

a Opposite side, angle $40°$, hypotenuse $10$: opp $=10\sin 40°=$

b Adjacent side, angle $28°$, hypotenuse $15$: adj $=15\cos 28°=$

c Opposite side, angle $55°$, adjacent $9$: opp $=9\tan 55°=$

Section 6 — Finding an angle (§6B)

7. Find $\theta$ to 1 d.p.

a opposite $7$, hypotenuse $12$: $\theta=\sin^{-1}\!\big(\tfrac{7}{12}\big)=$

b adjacent $6$, hypotenuse $11$: $\theta=\cos^{-1}\!\big(\tfrac{6}{11}\big)=$

c opposite $4$, adjacent $9$: $\theta=\tan^{-1}\!\big(\tfrac{4}{9}\big)=$

Section 7 — Elevation & depression (§6C)

8. Mr Wong stands $20$ m from the base of a tree. His eyes are $1.6$ m above the ground, and the angle of elevation to the top of the tree is $35°$. Find the height of the tree to 1 d.p. (Hint: tree height $=1.6+20\tan 35°$.) Height $=$

Section 8 — Bearings (§6D)

9. A ship sails $15$ km from port $P$ on a bearing of $040°$T. Find how far north and how far east of $P$ it now is (2 d.p.).

a north $=15\cos 40°=$

b east $=15\sin 40°=$

Challenge — 3D (§6E)

10. A closed rectangular box has base $6$ cm by $8$ cm and height $5$ cm. (a) Find the length of the longest straight rod that fits inside (the space diagonal), exact and to 2 d.p. (b) Find the angle this diagonal makes with the base, to 1 d.p. (Hint: the base diagonal is $\sqrt{6^{2}+8^{2}}=10$.)