IB Maths AI HL 3 Geometry and Trigonometry Questions

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

Exam points

Solve spatial geometry problems with coordinates, vectors, distances, angles, areas and volumes.
Apply triangle/radian trigonometry, identities and finite-interval equations to exact and applied geometric situations.
Use loci, perpendicular bisectors, Voronoi regions and graph features to classify or locate geometric structures.
Use vectors, line equations, scalar/vector products and kinematic models to solve 2D/3D positions, angles, intersections and distances.
Use transformation, adjacency and transition matrices to model geometric or network changes and interpret matrix powers.
Apply graph algorithms to routes, spanning trees and weighted-network bounds, checking parity, connectivity and edge weights.

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.

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 ]

$\delta=0.440$

(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.

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 .

[ 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)

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

Find the cost per metre of installing this cable.

[ 2 ]

$21310 \div 129.150 \ldots$
$ 165

(c)

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

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

[ 1 ]

(d)

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 (d) — 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.

[ 5 ]

(i)

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

[ 3 ]

(ii)

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

[ 2 ]

Finding the sum of the weights of their edges

8239+8668+8778+8811+8833+9251+11539

$\text{total cost}=\$64\,119$

(e)

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 ]

(f)

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 (f) — 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 = $\$ 90577$

(g)

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 (g) — 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 = $\$ 76109$

[Maximum number: 27]

The following question compares the distance and direction between cities on a flat surface to the distance and direction between cities on a sphere.
Consider a model where the cities of Bogotá, Moscow, and Nairobi lie on a flat surface. In this model, Nairobi is 6000 km due south of Moscow and Bogotá is 12500 km due west of Nairobi, as shown in the following diagram.

(a)

Find the distance from Bogotá to Moscow.

[ 2 ]

attempt to use Pythagoras' theorem

\begin{aligned} & \sqrt{12500^{2}+6000^{2}} \\ & =13900(\mathrm{~km}) \end{aligned}

Note: Answers to more than 3 sf are not appropriate and should be penalised. A1 will be given only for 13900 km ( 3 sf ) or 14000 km (nearest 500 km ).

(b)

Find the bearing of Moscow from Bogotá. Give your answer in degrees.

[ 3 ]

attempt to use an appropriate inverse trig ratio
e.g. $\arctan \left(\frac{12500}{6000}\right)$ or $\arctan \left(\frac{6000}{12500}\right)$
identifying the correct angle
$=(0) 64.4^{\circ}$ OR $064^{\circ}(64.3589 \ldots)$

Note: The two method marks can be awarded independently. The first can come from any geometry leading to an inverse trigonometric ratio. The second requires recognizing which angle is the bearing, possibly on a diagram. 64.4 appearing in intermediate working is not sufficient.

(c)

In reality, these three cities lie on the curved surface of the Earth which will change the distances and directions found in part (a).

Now consider a curved model using a coordinate system (x, y, z) with its origin, O , at the centre of the Earth. The units of this system are thousands of kilometres and the Earth is modelled as a sphere with radius 6000 km . The North Pole, P, lies on the z-axis, and Nairobi, N, is modelled as being on the equator and lying on the y-axis.

Figure for Question (c) — IB Maths AI HL

P has position vector $\overrightarrow{\mathrm{OP}}=\boldsymbol{p}=\left(\begin{array}{l}0 \\ 0 \\ 6\end{array}\right)$ and N has position vector $\overrightarrow{\mathrm{ON}}=\boldsymbol{n}=\left(\begin{array}{l}0 \\ 6 \\ 0\end{array}\right)$ .

[ 4 ]

(i)

Use the scalar product to find the angle between p and n.

[ 2 ]

$\quad \boldsymbol{p} . \boldsymbol{n}=0$
so the angle is $90^{\circ}$ OR $\frac{\pi}{2}$ rads

Note: Award M0A0 for an answer of 90 degrees without any mention that the scalar product is zero.

(ii)

Show that the distance between P and N along the arc from P to N is $3000 \pi \mathrm{~km}$ .

[ 2 ]

attempt to use arc length formula $s=r \theta$ OR $\frac{90^{\circ}}{360}$ of circumference
$=6000\left(\frac{\pi}{2}\right)$ OR $\frac{1}{4} \times 2 \pi \times 6000$ $=3 \pi$ thousand $\mathrm{km} \quad$ OR $\quad 3000 \pi \mathrm{~km}$

(d)

Point A, which is also on the equator, has position vector $\boldsymbol{a}=\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)$ as shown in the
following diagram.

P, N and A , and the arcs connecting them, form a spherical triangle.
The angle at vertex A is defined as the angle between the vectors $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ .

[ 5 ]

(i)

Find the vector $\boldsymbol{a} \times \boldsymbol{p}$ .

[ 2 ]

attempt to find cross product (e.g. correct except for sign error)
$\boldsymbol{a} \times \boldsymbol{p}=\left(\begin{array}{c}0 \\ -36 \\ 0\end{array}\right)$

(ii)

Show that the angle at vertex A in the spherical triangle is $90^{\circ}$ .

Moscow, M , has position vector $\overrightarrow{\mathrm{OM}}=\boldsymbol{m}=\left(\begin{array}{c}0 \\ 6 \cos \theta \\ 6 \sin \theta\end{array}\right)$ , as shown in the following diagram.

Figure for Question (ii) — IB Maths AI HL

The shortest distance between two points on the sphere lies along an arc of a circle on the sphere with centre O . In this model the shortest distance from Moscow to Nairobi is 6000 km .

[ 3 ]

METHOD 1
$\boldsymbol{a} \times \boldsymbol{n}=\left(\begin{array}{c}0 \\ 0 \\ 36\end{array}\right)$
choosing to use scalar product for their $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ $(\boldsymbol{a} \times \boldsymbol{p}) \cdot(\boldsymbol{a} \times \boldsymbol{n})=\mathbf{0}$

Note: The final A1 can be awarded for seeing the dot product equals zero without incorrect working. Follow through from (i) is permitted and if they explicitly state that they are using $\boldsymbol{a} \times \boldsymbol{n}$ then allow internal follow through.
so angle is $90^{\circ}$
AG

METHOD 2
$\boldsymbol{a} \times \boldsymbol{n}=\left(\begin{array}{c}0 \\ 0 \\ 36\end{array}\right)$
attempt to use cross product for their $\boldsymbol{a} \times \boldsymbol{p}$ and $\boldsymbol{a} \times \boldsymbol{n}$ . This must include both an attempt to find the cross product and knowledge that $\sin \theta=\frac{|x \times y|}{|x||y|}$ $(\boldsymbol{a} \times \boldsymbol{n}) \times(\boldsymbol{a} \times \boldsymbol{p})=\left(\begin{array}{c}1296 \\ 0 \\ 0\end{array}\right) \quad$ OR $(\boldsymbol{a} \times \boldsymbol{p}) \times(\boldsymbol{a} \times \boldsymbol{n})=\left(\begin{array}{c}-1296 \\ 0 \\ 0\end{array}\right)$ $\sin \theta=\frac{1296}{36 \times 36}=1$
so angle is $90^{\circ}$

(e)

Show that $\theta=57.3^{\circ}$ , correct to three significant figures.

Bogotá, B , is west of Nairobi and has position vector $\overrightarrow{\mathrm{OB}}=\boldsymbol{b}=\left(\begin{array}{c}6 \sin 120^{\circ} \\ 6 \cos 120^{\circ} \\ 0\end{array}\right)$ .

[ 2 ]

attempt to use $\frac{\theta}{360} \times 2 \pi r=6$ OR $s=r \theta$

\frac{\theta}{360} \times 2 \pi \times 6=6 \quad \text { OR } \quad 6=6 \theta \Rightarrow \theta=1 \text { radian }

$\theta=57.2957 \ldots{ }^{\circ}$

Note: The fourth significant figure must be seen somewhere to award the A1.

$\theta=57.3^{\circ}$ to 3 significant figures

(f)

Find the shortest distance from Bogotá to Moscow on the sphere.

The bearing from B to M is defined as the angle at vertex B in the spherical triangle containing B, M and P . It is given that $\boldsymbol{b} \times \boldsymbol{p}=\left(\begin{array}{c}36 \cos 120^{\circ} \\ -36 \sin 120^{\circ} \\ 0\end{array}\right)$ .

[ 5 ]

(e) METHOD 1

EITHER (Scalar product)
attempt to use scalar product to find BÔM $=\arccos \left(\frac{\boldsymbol{b} . \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}\right)$

Note: This may be written as $\cos (\mathrm{BOM})=\frac{\boldsymbol{b} \boldsymbol{.} \boldsymbol{m}}{|\boldsymbol{b}||\boldsymbol{m}|}$ .

=\arccos \left(\frac{\left(6 \cos 120^{\circ}\right)\left(6 \cos 57.3^{\circ}\right)}{6^{2}}\right)

Note: This A1 can be awarded for $\cos \theta=\frac{6 \cos 120^{\circ} \times 6 \cos 57.3^{\circ}}{6^{2}}$ .

=106^{\circ}\left(=105.673 \ldots{ }^{\circ}\right) \text { OR } 1.84 \text { radians }(1.84434 \ldots)

OR (Cosine Rule)

\begin{aligned} & |\mathrm{BM}|=\sqrt{(6 \sin 120-0)^{2}+(6 \cos 120-6 \sin 57.3)^{2}+(0-6 \sin 57.3)^{2}} \\ & =9.56(9.56299 \ldots) \text { OR } 9560 \mathrm{~km} \end{aligned}

attempt to use the cosine rule to find angle BOM
$|\mathrm{BM}|^{2}=6^{2}+6^{2}-2 \times 6 \times 6 \times \cos \mathrm{BOM}$

\mathrm{BOM}=106^{\circ}\left(=105.673 \ldots{ }^{\circ}\right) \text { OR } 1.84 \text { radians }(1.84434 \ldots)

THEN

\begin{aligned} & \text { attempt to use } \frac{\theta}{360} \times 2 \pi r \text { OR } s=r \theta \\ & =\frac{105.671^{\circ}}{360} \times 2 \pi \times 6 \\ & =11.1(11.0660 \ldots \text { thousand } \mathrm{km}) \end{aligned}

Note: Accept an answer of 11100 OR $11066.0 \ldots$ without units (i.e. "km" is implicit).

METHOD 2 (Cross product)

\text { attempt to use cross product to find BÔM }=\arcsin \frac{|\boldsymbol{b} \times \boldsymbol{m}|}{|\boldsymbol{b}||\boldsymbol{m}|}

Note: This may be written as $\sin$ BÔM $=\frac{|\boldsymbol{b} \times \boldsymbol{m}|}{|\boldsymbol{b}||\boldsymbol{m}|}$ .

\begin{aligned} & =\frac{\left|\left(\begin{array}{c} 36 \cos 120 \sin \theta \\ -36 \sin 120 \sin \theta \\ 36 \sin 120 \cos \theta \end{array}\right)\right|}{6 \times 6} \\ & =\frac{\sqrt{\sin ^{2} \theta+3}}{2}=0.962818 \ldots \end{aligned}

\begin{aligned} \text { BÔM } & =180-\arcsin 0.962818 \ldots \\ & =106^{\circ}(105.673 \ldots, 1.844346 \mathrm{rads}) \end{aligned}

\text { attempt to use } \frac{\theta}{360} \times 2 \pi r \quad \text { OR } \quad s=r \theta

=\frac{105.671}{360} \times 2 \pi \times 6

=11.1 (11.0660 thousand km)

Note: Award at most (M1)(A1)(A0)(M1)A0 for an answer of 7.78 thousand km from finding the principal root of the arcsin.

(g)

Using the method from part (c), find the bearing from Bogotá to Moscow.
Disclaimer:
Content used in IB assessments is taken from authentic, third-party sources. The views expressed within them belong to their individual authors and/or publishers and do not necessarily reflect the views of the IB.

[ 6 ]

attempt to calculate $\boldsymbol{b} \times \boldsymbol{m}$ $\boldsymbol{b} \times \boldsymbol{m}=\left(\begin{array}{c}36 \cos 120^{\circ} \sin 57.3^{\circ} \\ -36 \sin 120^{\circ} \sin 57.3^{\circ} \\ 36 \sin 120^{\circ} \cos 57.3^{\circ}\end{array}\right)\left(=\left(\begin{array}{c}-15.1464 \ldots \\ -26.2344 \ldots \\ 16.8449 \ldots\end{array}\right)\right)$
attempt to use scalar product to find angle between vectors
$(\cos \theta=)\left(\frac{(\boldsymbol{b} \times \boldsymbol{m}) \cdot(\boldsymbol{b} \times \boldsymbol{p})}{|\boldsymbol{b} \times \boldsymbol{m} \| \boldsymbol{b} \times \boldsymbol{p}|}\right)$ $(\cos \theta=)\left(\frac{1090.54 \ldots}{34.6614 \ldots \times 36}\right)$

Note: Award (A1) for numerator, (A1) for denominator.
bearing is $(0) 29.1^{\circ}\left(29.0770 \ldots .^{\circ}\right)$

Note: Do not penalise absence of degree symbol.
Award full marks for answer in radians =0.507.
Award full marks for $029^{\circ}$ .
Special Case: If a candidate calculates $\boldsymbol{m} \times \boldsymbol{b}$ instead of $\boldsymbol{b} \times \boldsymbol{m}$ then you can award at most M1A0M1A1A1A0.

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (i)

Question (ii)

Question (e)

Question (f)

Question (g)

Question 4

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question (i)

Question (ii)

Question (e)

Question (f)

Question (g)