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IB Maths AI SL – Topic 3: Geometry and Trigonometry

Right-angled and non-right-angled trigonometry, Voronoi diagrams, 3D solids.

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12 questions  ·  70 marks  ·  answers included  ·  IB AI SL
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This worksheet covers:

  • Distance and midpoint in 3D
  • Volume and surface area of solids
  • Right-angled and non-right-angled trigonometry (sine and cosine rules)
  • Arcs and sectors
  • Bearings
  • Voronoi diagrams

A GDC is allowed throughout; allow about 75 minutes. Give answers to 3 significant figures unless exact.

Questions

Section A – Skills

Short questions. GDC allowed.
1

The points $A(1, 2, 3)$ and $B(5, -2, 7)$ are two vertices of a solid.(a) Find the distance $AB$.(b) Find the coordinates of the midpoint of $AB$.

Show answer
(a) $\sqrt{16 + 16 + 16} = \sqrt{48} = 6.93$(b) $(3,\ 0,\ 5)$
[4]
2

A ladder of length $6$ m leans against a vertical wall, making an angle of $70^\circ$ with the horizontal ground.(a) Find how far up the wall the ladder reaches.(b) Find the distance of the foot of the ladder from the wall.

Show answer
(a) $6\sin 70^\circ = 5.64$ m(b) $6\cos 70^\circ = 2.05$ m
[4]
3

A solid toy consists of a cone of radius $4$ cm and height $9$ cm attached to a hemisphere of radius $4$ cm.(a) Find the volume of the cone.(b) Find the total volume of the toy.

Show answer
(a) $\frac{1}{3}\pi (4^2)(9) = 48\pi = 151$ cm$^3$(b) $48\pi + \frac{2}{3}\pi(4^3) = 48\pi + 134.0 = 285$ cm$^3$
[4]
4

In triangle $ABC$, angle $A = 42^\circ$, angle $B = 67^\circ$ and $a = BC = 8.5$ cm.(a) Find the length of $AC$.(b) Find the length of $AB$.(c) Find the area of the triangle.

Show answer
(a) $b = \dfrac{8.5\sin 67^\circ}{\sin 42^\circ} = 11.7$ cm(b) $C = 71^\circ$, $c = \dfrac{8.5\sin 71^\circ}{\sin 42^\circ} = 12.0$ cm(c) $\frac{1}{2}(8.5)(11.69)\sin 71^\circ = 47.0$ cm$^2$
[6]
5

A sector of a circle has radius $12$ cm and angle $150^\circ$.(a) Find the length of the arc.(b) Find the area of the sector.

Show answer
(a) $\dfrac{150}{360}\times 2\pi(12) = 10\pi = 31.4$ cm(b) $\dfrac{150}{360}\times \pi(12^2) = 60\pi = 188$ cm$^2$
[4]

Section B – Exam-style

Show full method. GDC allowed.
6

In triangle $PQR$, $PQ = 7$ cm, $QR = 9$ cm and angle $PQR = 110^\circ$.(a) Find the length of $PR$.(b) Find the size of angle $QRP$.

Show answer
(a) $PR^2 = 49 + 81 – 2(7)(9)\cos 110^\circ$, $PR = 13.2$ cm(b) $\cos R = \dfrac{81 + 13.16^2 – 49}{2(9)(13.16)}$, $R = 30.0^\circ$
[5]
7

A ship leaves port $O$ and sails $12$ km on a bearing of $060^\circ$ to point $A$. It then sails $20$ km on a bearing of $150^\circ$ to point $B$.(a) Show that angle $OAB = 90^\circ$.(b) Find the distance $OB$.(c) Find the bearing of $B$ from $O$.

Show answer
(a) At $A$ the bearing back to $O$ is $060^\circ + 180^\circ = 240^\circ$ and the bearing of $B$ is $150^\circ$; angle $OAB = 240^\circ – 150^\circ = 90^\circ$(b) $OB = \sqrt{12^2 + 20^2} = 23.3$ km(c) Angle $AOB = \tan^{-1}(20/12) = 59.0^\circ$; bearing $= 060 + 59.0 = 119^\circ$
[6]
8

Three mobile-phone masts are located at $A(2, 2)$, $B(8, 4)$ and $C(4, 8)$, with coordinates in km. A Voronoi diagram is to be drawn so that each phone connects to its nearest mast.(a) Find the equation of the perpendicular bisector of $AB$.(b) Show that the perpendicular bisector of $AC$ has equation $x + 3y = 18$.(c) Find the coordinates of the vertex of the Voronoi diagram.(d) A phone is at $(5, 3)$. Determine which mast it connects to.

Show answer
(a) Midpoint $(5, 3)$, gradient $-3$: $y = -3x + 18$ (or $3x + y = 18$)(b) Midpoint $(3, 5)$, $AC$ gradient $3$ so bisector gradient $-\frac{1}{3}$: $y – 5 = -\frac{1}{3}(x – 3) \Rightarrow x + 3y = 18$(c) $(4.5,\ 4.5)$(d) $(5,3)$ lies on the bisector of $AB$ so $A$ and $B$ are equidistant ($\sqrt{10}$); distance to $C$ is $\sqrt{26}$. Connects to $A$ or $B$ (equidistant), not $C$.
[8]
9

A cuboid has a base $8$ cm by $6$ cm and height $5$ cm. Vertex $A$ is a bottom corner and vertex $G$ is the top corner diagonally opposite.(a) Find the length of the base diagonal.(b) Find the length $AG$.(c) Find the angle that $AG$ makes with the base.

Show answer
(a) $\sqrt{64 + 36} = 10$ cm(b) $\sqrt{100 + 25} = \sqrt{125} = 11.2$ cm(c) $\tan^{-1}(5/10) = 26.6^\circ$
[5]
10

From a point $P$ on level ground, the angle of elevation of the top of a vertical tower is $35^\circ$. From a point $Q$, $40$ m closer to the tower and in line with $P$ and the base, the angle of elevation is $52^\circ$.(a) Let the distance from $Q$ to the base be $x$ m. Write down two expressions for the height $h$ of the tower in terms of $x$.(b) Hence find $x$ and the height of the tower.

Show answer
(a) $h = x\tan 52^\circ$ and $h = (x + 40)\tan 35^\circ$(b) $x = \dfrac{40\tan 35^\circ}{\tan 52^\circ – \tan 35^\circ} = 48.3$ m; $h = 61.8$ m
[6]

Section C – Extended

Longer structured questions in context.
11

A grain silo consists of a cylinder of radius $1.2$ m and height $3.5$ m with a conical roof of the same radius and height $0.9$ m.(a) Find the volume of the cylinder, in m$^3$.(b) Find the slant height of the cone.(c) The curved surface of the cylinder and the curved surface of the cone are to be painted. Find the total area to be painted.(d) Paint costs $\$4.80$ per square metre. Find the cost of painting the silo, to the nearest dollar.(e) Find the total volume of the silo, in litres ($1\ \text{m}^3 = 1000$ litres).

Show answer
(a) $\pi(1.2^2)(3.5) = 15.8$ m$^3$(b) $\sqrt{1.2^2 + 0.9^2} = 1.5$ m(c) $2\pi(1.2)(3.5) + \pi(1.2)(1.5) = 26.39 + 5.65 = 32.0$ m$^2$(d) $32.04 \times 4.80 = \$154$(e) $15.83 + 1.357 = 17.19$ m$^3 = 17\,200$ litres
[9]
12

A triangular field $ABC$ has $AB = 120$ m, $AC = 95$ m and angle $BAC = 58^\circ$.(a) Find the length of $BC$.(b) Find the area of the field, in hectares ($1$ ha $= 10\,000$ m$^2$).(c) Find the size of angle $ABC$.(d) The farmer fences the whole perimeter at a cost of $\$12.50$ per metre. Find the total cost.(e) A water trough is to be placed at the point on $BC$ nearest to $A$. Find the distance from $A$ to the trough.

Show answer
(a) $BC^2 = 120^2 + 95^2 – 2(120)(95)\cos 58^\circ$, $BC = 107$ m ($106.5$)(b) $\frac{1}{2}(120)(95)\sin 58^\circ = 4834$ m$^2 = 0.483$ ha(c) $\sin B = \dfrac{95\sin 58^\circ}{106.5}$, $B = 49.2^\circ$(d) Perimeter $= 321.5$ m; cost $= \$4019$ ($\$4018.78$)(e) Perpendicular height from $A$: $120\sin 49.15^\circ = 90.8$ m (or $2 \times 4834 / 106.5$)
[9]

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