IB Maths AA HL 3.1 Geometry and Trigonometry Sl Content Questions

Practise IB Mathematics AA HL 3.1 by combining vectors, trigonometric identities, loci and advanced geometry with exact reasoning.

Syllabus
First assessment 2021
Course
Mathematics: analysis and approaches HL
Level
HL

Exam points

  • Solve 3D length, midpoint, volume or surface-area problems by combining coordinates, solid geometry and right-triangle distances.
  • Apply right/non-right triangle trigonometry, Pythagoras, bearings and elevation/depression to distances, angles, areas or travel models.
  • Convert and use radians to calculate arc lengths, sector/segment areas and related geometric parameters.
  • Use the unit circle, exact values, identities and double-angle relationships to transform or simplify trigonometric expressions.
  • Graph and analyse sine, cosine or tangent functions, recover amplitude/period/phase parameters, and solve finite-interval equations or periodic models.

Question 1

[Maximum number: 6]

The points A and B lie on a circle, with centre O and radius 19.5 cm , such that BOO^=210∘\mathrm{BO} \widehat{\mathrm{O}}=210^{\circ}.
A piece of paper is cut into the shape of the sector BOA .
A hollow cone with no base is constructed from the sector by joining the points A and B . The sector forms the curved surface of the cone.
This is shown in the following diagrams.

Figure for Question 1 — IB Maths AA HL

Find

Question (a)

(a)

the area of the sector BOA ;

[ 3 ]

Question (b)

(b)

the radius of the cone.

[ 3 ]

Question 2

[Maximum number: 5]

Consider the triangle PQR where QPR^=30∘,PQ=(x+2)cm\mathrm{Q} \hat{\mathrm{PR}}=30^{\circ}, \mathrm{PQ}=(x+2) \mathrm{cm} and PR=(5−x)2 cm\mathrm{PR}=(5-x)^{2} \mathrm{~cm}, where -2<x<5.

Question (a)

(a)

Show that the area, A cm2A \mathrm{~cm}^{2}, of the triangle is given by A=14(x3−8x2+5x+50)A=\frac{1}{4}\left(x^{3}-8 x^{2}+5 x+50\right).

[ 2 ]

Question (b)

(b)

Find QR when the area of triangle PQR is a maximum.

[ 3 ]

Question 3

[Maximum number: 9]

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.

Figure for Question 3 — IB Maths AA HL

Question (a)

(a)

Find the distance from Bogotá to Moscow.

[ 2 ]

Question (b)

(b)

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

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 (b) — IB Maths AA HL

P has position vector OP→=p=(006)\overrightarrow{\mathrm{OP}}=\boldsymbol{p}=\left(\begin{array}{l}0 \\ 0 \\ 6\end{array}\right) and N has position vector ON→=n=(060)\overrightarrow{\mathrm{ON}}=\boldsymbol{n}=\left(\begin{array}{l}0 \\ 6 \\ 0\end{array}\right).

[ 3 ]

Question (c)

(c)

Show that the distance between P and N along the arc from P to N is 3000π km3000 \pi \mathrm{~km}.

Point A, which is also on the equator, has position vector a=(600)\boldsymbol{a}=\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right) as shown in the following
diagram.

Figure for Question (c) — IB Maths AA HL

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 a×p\boldsymbol{a} \times \boldsymbol{p} and a×n\boldsymbol{a} \times \boldsymbol{n}.

[ 2 ]

Question (d)

(d)

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

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

[ 2 ]
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