IB Maths AI HL 3 Geometry and Trigonometry

Practise IB Maths AI HL geometry and trigonometry through shared-core and HL diagrams, bearings, vectors, radians and technology-supported solutions.

Covered syllabus points

3.1 Geometry and trigonometry - SL content 3.2 Geometry and trigonometry - AHL content

[Maximum number: 7]

Kailash manufactures drink containers in the shape of a cuboid. The container has a square top and a square base of length, $l\,\mathrm{cm}$ . Its height, $d\,\mathrm{cm}$ , is three times the length of the base.

diagram not to scale

(a)

Calculate the total external surface area of the container.

[ 3 ]

Finding the area of a square and the area of a non-square rectangle: $5\times5$ and $5\times15$ .
$(5\times5+5\times15+5\times15)\times2$ $=350\,\mathrm{cm^2}$
Note: Units are required.

(b)

To reduce environmental impact, Kailash is trying to minimize the amount of material needed for the production of the $375\,\mathrm{cm^3}$ container.
He is willing to change the shape to a cylinder with radius $r\,\mathrm{cm}$ , and height $h\,\mathrm{cm}$ , as shown below.

Figure for Question (b) — IB Maths AI HL

The cylindrical container of drink must also hold $375\,\mathrm{cm^3}$ .
Find an expression for the height, h, of the container in terms of r.

[ 2 ]

$\pi r^2h=375$ $h=\frac{375}{\pi r^2}$
Note: Award at most A0A1 for a final answer $h=\frac{V}{\pi r^2}$ .

(c)

Let the total external surface area be $A\,\mathrm{cm^2}$ .
Show that $A=2\pi r^2+\frac{750}{r}$ .

[ 2 ]

Attempt to substitute their h into the curved surface area $2\pi rh$ .
$A=2\pi r^2+2\pi r\left(\frac{375}{\pi r^2}\right)$ $A=2\pi r^2+\frac{750}{r}$

[Maximum number: 9]

A suitable site for the landing of a spacecraft on the planet Mars is identified at a point, A. The shortest time from sunrise to sunset at point A must be found.
Radians should be used throughout this question. All values given in the question should be treated as exact.
Mars completes a full orbit of the Sun in 669 Martian days, which is one Martian year.

Figure for Question 3 Geometry and trigonometry question 2 — IB Maths AI HL

On day t, where $t \in \mathbb{Z}$ , the length of time, in hours, from the start of the Martian day until sunrise at point A can be modelled by a function, R(t), where

R(t)=a \sin (b t)+c, t \in \mathbb{R} .

The graph of R is shown for one Martian year.

(a)

Find the angle through which Mars rotates on its axis each hour.

The time of sunrise on Mars depends on the angle, $\delta$ , at which it tilts towards the Sun. During a Martian year, $\delta$ varies from -0.440 to 0.440 radians.

The angle, $\omega$ , through which Mars rotates on its axis from the start of a Martian day to the moment of sunrise, at point A , is given by $\cos \omega=0.839 \tan \delta, 0 \leq \omega \leq \pi$ .

[ 3 ]

length of day $=24 \frac{2}{3}$ hours

Note: Award A1 for $\frac{2}{3}, 0.666 \ldots, 0 . \overline{6}$ or 0.667 .

\frac{2 \pi}{24 \frac{2}{3}}

Note: Accept $\left(\frac{360}{24 \frac{2}{3}}\right)$ .

=0.255 \text { radians }\left(=0.254723 \ldots, \frac{3 \pi}{37}, 14.5945 \ldots{ }^{\circ}\right)

(b)

Show that the maximum value of $\omega=1.98$ , correct to three significant figures.

[ 3 ]

substitution of either value of $\delta$ into equation

Note: Both (M1) lines may be seen in either part (c)(i) or part (c)(ii).

\cos \omega=0.839 \tan (-0.440)

\omega=1.97684 \ldots

\approx 1.98

Note: For substitution of 1.98 award MOAO.

(c)

Find the minimum value of $\omega$ .

[ 1 ]

$\delta=0.440$

A1

(d)

Hence show that a=1.6, correct to two significant figures.

[ 2 ]

$a=\frac{7.76075 \ldots-4.57258 \ldots}{2}$ $\approx 1.59408 \ldots$

Note: Award M1 for substituting their values into a correct expression.

Award A1 for a correct value of a from their expression which has at least 3 significant figures and rounds correctly to 1.6 .
$\approx 1.6$ (correct to 2 sf )

[Maximum number: 23]

This question compares possible designs for a new computer network between multiple school buildings, and whether they meet specific requirements.
A school's administration team decides to install new fibre-optic internet cables underground. The school has eight buildings that need to be connected by these cables. A map of the school is shown below, with the internet access point of each building labelled A-H.

Figure for Question 3 Geometry and trigonometry question 3 — IB Maths AI HL

Jonas is planning where to install the underground cables. He begins by determining the distances, in metres, between the underground access points in each of the buildings.

He finds $\mathrm{AD}=89.2 \mathrm{~m}, \mathrm{DF}=104.9 \mathrm{~m}$ and $\mathrm{A} \hat{\mathrm{DF}}=83^{\circ}$ .

(a)

Find AF .

The cost for installing the cable directly between A and F is $\$ 21310$ .

[ 3 ]

$\quad \mathrm{AF}^{2}=89.2^{2}+104.9^{2}-2(89.2)(104.9) \cos 83$
(M1)(A1)
Note: Award (M1) for substitution into the cosine rule and (A1) for correct substitution.

\mathrm{AF}=129 \mathrm{~m} \quad(129.150 \ldots)

(b)

Find the cost per metre of installing this cable.

Jonas estimates that it will cost $\$ 110$ per metre to install the cables between all the other buildings.

[ 2 ]

$21310 \div 129.150 \ldots$
$ 165

(c)

State why the cost for installing the cable between A and F would be higher than between the other buildings.

Jonas creates the following graph, S, using the cost of installing the cables between two buildings as the weight of each edge.

Figure for Question (c) — IB Maths AI HL

The computer network could be designed such that each building is directly connected to at least one other building and hence all buildings are indirectly connected.

[ 1 ]

(d)

By using Kruskal's algorithm, find the minimum spanning tree for S, showing clearly the order in which edges are added.

[ 3 ]

(e)

Hence find the minimum installation cost for the cables that would allow all the buildings to be part of the computer network.

[ 2 ]

(f)

The computer network fails if any part of it becomes unreachable from any other part. To help protect the network from failing, every building could be connected to at least two other buildings. In this way if one connection breaks, the building is still part of the computer network. Jonas can achieve this by finding a Hamiltonian cycle within the graph.

State why a path that forms a Hamiltonian cycle does not always form an Eulerian circuit.

[ 1 ]

(g)

Starting at D, use the nearest neighbour algorithm to find the upper bound for the installation cost of a computer network in the form of a Hamiltonian cycle.

Note: Although the graph is not complete, in this instance it is not necessary to form a table of least distances.

[ 5 ]

edges (or weights) are chosen in the order

DG (8668)
GH (9603)
HE (9251)
EC (8239)
CB (13 156)
BA (8811)
AF (21 310)
FD (11 539)
A1A1A1

Figure for Answer to Question (g) — IB Maths AI HL

Note: Award A1 for the first two edges chosen in the correct order. Award A1A1 for the first five edges chosen in the correct order. Award A1A1A1 for all eight edges chosen in the correct order. Accept a diagram as an answer, provided the order of edges is communicated.
finding the sum of the weights of their edges
8668+9603+9251+8239+13156+8811+21310+11539
upper bound = $\$ 90577$

(h)

By deleting D, use the deleted vertex algorithm to find the lower bound for the installation cost of the cycle.

[ 6 ]

attempt to find MST after deleting vertex D
these edges (or weights) (in any order)
CE (8239)
AB (8811)
EH (9251)
GH (9603)
BE (10 153)
FG (12606)

Note: Prim's or Kruskal's algorithm could be used at this stage.
reconnect D to MST with two different edges

DG (8668)
BD (8778)

Note: This A1 is independent of the first A mark and can be awarded if both DG and BD are chosen to reconnect D to the MST, even if the MST is incorrect.

Figure for Answer to Question (h) — IB Maths AI HL

finding the sum of the weights of their edges
8239+8811+9251+9603+10153+12606+8668+8778
Note: For candidates with an incorrect MST or no MST, the weights of at least seven of the edges being summed (two of which must connect to D ) must be shown to award this (M1).
lower bound = $\$ 76109$

All question bank results loaded

Covered syllabus points

3 Geometry and trigonometry question 1

Question (a)

Question (b)

Question (c)

3 Geometry and trigonometry question 2

Question (a)

Question (b)

Question (c)

Question (d)

3 Geometry and trigonometry question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)

Question (h)