Home › Free Resources › Maths Worksheets › IB Maths AA SL – Topic 3: Geometry and Trigonometry
Free Maths Worksheet

IB Maths AA SL – Topic 3: Geometry and Trigonometry

Radians, trigonometric identities and equations, 3D geometry, circle geometry.

WhatsApp Mr. AhmadBook a trial lesson (50% off)Free worksheets
12 questions  ·  62 marks  ·  answers included  ·  IB AA SL
Download the PDFAsk about a question

This worksheet covers:

  • Radians
  • Arcs and sectors
  • The sine and cosine rules
  • Trigonometric identities and equations
  • 3D geometry

Section A is non-calculator; Sections B and C allow a GDC. Allow about 75 minutes. Give answers to 3 significant figures unless exact values are requested.

Questions

Section A – Skills

Non-calculator. Exact answers required.
1

(a) Convert $150^\circ$ to radians, giving your answer in terms of $\pi$.(b) Convert $\dfrac{7\pi}{4}$ radians to degrees.(c) Write down the exact value of $\cos\left(\dfrac{5\pi}{6}\right)$.

Show answer
(a) $\dfrac{5\pi}{6}$(b) $315^\circ$(c) $-\dfrac{\sqrt{3}}{2}$
[4]
2

A sector of a circle has radius $6\,\text{cm}$ and angle $1.2$ radians at the centre.(a) Find the length of the arc.(b) Find the area of the sector.(c) Find the perimeter of the sector.

Show answer
(a) $7.2\,\text{cm}$(b) $21.6\,\text{cm}^2$(c) $19.2\,\text{cm}$
[4]
3

Given that $\tan\theta = -2$ and $\dfrac{\pi}{2} < \theta < \pi$, find the exact values of(a) $\sin\theta$ and $\cos\theta$,(b) $\sin 2\theta$,(c) $\cos 2\theta$.

Show answer
(a) $\sin\theta = \dfrac{2}{\sqrt{5}}$, $\cos\theta = -\dfrac{1}{\sqrt{5}}$(b) $\sin 2\theta = -\dfrac{4}{5}$(c) $\cos 2\theta = -\dfrac{3}{5}$
[5]
4

Show that $\dfrac{1 – \cos 2x}{\sin 2x} \equiv \tan x$, for $\sin 2x eq 0$.

Show answer
$1-\cos 2x = 2\sin^2 x$ and $\sin 2x = 2\sin x\cos x$, so the quotient is $\dfrac{2\sin^2 x}{2\sin x\cos x} = \dfrac{\sin x}{\cos x} = \tan x$.
[3]
5

Solve $2\cos^2 x + 3\sin x = 3$ for $0 \le x \le 2\pi$. Give exact answers.

Show answer
Using $\cos^2 x = 1 – \sin^2 x$: $2\sin^2 x – 3\sin x + 1 = 0$, so $\sin x = 1$ or $\sin x = \dfrac{1}{2}$.$x = \dfrac{\pi}{6},\ \dfrac{\pi}{2},\ \dfrac{5\pi}{6}$
[5]

Section B – Exam-style

Calculator allowed.
6

In triangle $ABC$, $AB = 8\,\text{cm}$, $AC = 11\,\text{cm}$ and $\hat{A} = 52^\circ$.(a) Find the length of $BC$.(b) Find the area of the triangle.

Show answer
(a) $BC^2 = 8^2 + 11^2 – 2(8)(11)\cos 52^\circ$, so $BC = 8.75\,\text{cm}$(b) Area $= \tfrac{1}{2}(8)(11)\sin 52^\circ = 34.7\,\text{cm}^2$
[5]
7

In triangle $PQR$, $PQ = 7\,\text{cm}$, $\hat{P} = 65^\circ$ and $\hat{R} = 40^\circ$.(a) Find $\hat{Q}$.(b) Find the length of $QR$.

Show answer
(a) $\hat{Q} = 75^\circ$(b) $\dfrac{QR}{\sin 65^\circ} = \dfrac{7}{\sin 40^\circ}$, so $QR = 9.87\,\text{cm}$
[4]
8

Solve $\sin\left(2x – \dfrac{\pi}{3}\right) = \dfrac{1}{2}$ for $0 \le x \le \pi$.

Show answer
$2x – \dfrac{\pi}{3} = \dfrac{\pi}{6}$ or $\dfrac{5\pi}{6}$ (the range $-\dfrac{\pi}{3} \le 2x – \dfrac{\pi}{3} \le \dfrac{5\pi}{3}$ gives no others).$x = \dfrac{\pi}{4},\ \dfrac{7\pi}{12}$
[5]
9

A cuboid $ABCDEFGH$ has base $ABCD$ with $AB = 6\,\text{cm}$ and $BC = 4\,\text{cm}$, and vertical height $AE = 3\,\text{cm}$. Vertex $G$ is vertically above $C$.(a) Find the length of the base diagonal $AC$.(b) Find the length of the space diagonal $AG$.(c) Find the angle that $AG$ makes with the base $ABCD$.

Show answer
(a) $AC = \sqrt{52} = 7.21\,\text{cm}$(b) $AG = \sqrt{61} = 7.81\,\text{cm}$(c) $\tan\theta = \dfrac{3}{\sqrt{52}}$, so $\theta = 22.6^\circ$
[6]
10

A solid right cone has base radius $5\,\text{cm}$ and slant height $13\,\text{cm}$.(a) Show that the vertical height of the cone is $12\,\text{cm}$.(b) Find the volume of the cone.(c) Find the total surface area of the cone.

Show answer
(a) $h = \sqrt{13^2 – 5^2} = \sqrt{144} = 12\,\text{cm}$(b) $V = \tfrac{1}{3}\pi(5^2)(12) = 100\pi = 314\,\text{cm}^3$(c) $A = \pi(5)(13) + \pi(5^2) = 90\pi = 283\,\text{cm}^2$
[6]

Section C – Challenge

Calculator allowed.
11

A chord $AB$ of length $10\,\text{cm}$ is drawn in a circle with centre $O$ and radius $7\,\text{cm}$.(a) Find the angle $A\hat{O}B$ in radians.(b) Find the area of the minor segment cut off by the chord.(c) Find the perimeter of the minor segment.

Show answer
(a) $\sin\left(\tfrac{\theta}{2}\right) = \dfrac{5}{7}$, so $\theta = 1.59$ radians(b) Area $= \tfrac{1}{2}(7^2)(\theta – \sin\theta) = 14.5\,\text{cm}^2$(c) Arc $= 7\theta = 11.1\,\text{cm}$, so perimeter $= 21.1\,\text{cm}$
[7]
12

In triangle $ABC$, $a = BC = 8\,\text{cm}$, $b = AC = 10\,\text{cm}$ and $\hat{A} = 50^\circ$.(a) Show that there are two possible triangles, and find both possible values of $\hat{B}$.(b) For each triangle, find the length of $AB$.(c) Find the difference between the areas of the two triangles.

Show answer
(a) $\sin B = \dfrac{10\sin 50^\circ}{8} = 0.958 < 1$; since $b > a$, both $B = 73.2^\circ$ and $B = 106.8^\circ$ give $A + B < 180^\circ$.(b) $C = 56.8^\circ$ gives $AB = 8.73\,\text{cm}$; $C = 23.2^\circ$ gives $AB = 4.12\,\text{cm}$(c) Areas $\tfrac{1}{2}(10)(c)\sin 50^\circ$: $33.5\,\text{cm}^2$ and $15.8\,\text{cm}^2$; difference $17.7\,\text{cm}^2$
[8]

Related worksheets

More worksheets in IB Mathematics: Analysis & Approaches:

IB Maths AA SL – Topic 1: Number and AlgebraIB Maths AA SL – Topic 2: FunctionsIB Maths AA SL – Topic 4: Statistics and ProbabilityIB Maths AA SL – Topic 5: CalculusIB Maths AA HL – Topic 1: Number and AlgebraIB Maths AA HL – Topic 2: FunctionsIB Maths AA HL – Topic 3: Geometry and TrigonometryIB Maths AA HL – Topic 4: Statistics and Probability

All maths worksheets  ·  Full resource library  ·  Maths tuition with Mr. Ahmad

Get new maths worksheets by email

One short email when new worksheets are added. No spam, unsubscribe any time.



    Book a trial lesson (50% off)

    Try a full first lesson with Mr. Ahmad at 50% off before you commit to anything. It is a real lesson, not a sales call: bring a topic the student is stuck on and we will work through it together.

    • Online from anywhere in the world, or in person across Dubai
    • You get honest feedback on where the student stands and what to fix first
    • No contract — if it is not the right fit, that is fine

    Prefer to talk? Message on WhatsApp and you will hear back within 24 hours.



      Stuck on this topic?

      Book a trial lesson with Mr. Ahmad at 50% off and bring this worksheet with you — online worldwide or in person in Dubai.

      WhatsApp nowBook a trial lesson (50% off)
      Scroll to Top