IB Mathematics AA HL 3.1 Geometry and Trigonometry Question Bank

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

Exam points

HL syllabus concepts and methods
HL extended reasoning

Covered syllabus points

SL 3.1—Three-dimensional geometry SL 3.2—Triangle trigonometry SL 3.3—Applications of trigonometry SL 3.4—Radian measure SL 3.5—Unit circle and exact values SL 3.6—Trigonometric identities SL 3.7—Trigonometric functions SL 3.8—Trigonometric equations

[Maximum number: 6]

The points A and B lie on a circle, with centre O and radius 19.5 cm , such that $\mathrm{BO} \widehat{\mathrm{O}}=210^{\circ}$ .
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 3.1 Geometry and trigonometry - SL content question 1 — IB Maths AA HL

Find

(a)

the area of the sector BOA ;

[ 3 ]

EITHER
$210^{\circ}=\frac{210 \pi}{180}\left(=\frac{7 \pi}{6}=3.66519 \ldots\right)$ radians
attempt to use radian formula for area of sector
area $=\frac{1}{2}(19.5)^{2}\left(\frac{7 \pi}{6}\right)$
OR
attempt to use degree formula for area of sector
area $=\frac{210}{360} \pi(19.5)^{2}$

THEN
area $=\frac{3549 \pi}{16}=696.844 \ldots$ $=697\left(=\frac{3549 \pi}{16}\right)\left(\mathrm{cm}^{2}\right)$

(b)

the radius of the cone.

[ 3 ]

EITHER
arc length $=19.5\left(\frac{7 \pi}{6}\right)$ OR $=\frac{210}{360}(2 \pi(19.5))\left(=\frac{91 \pi}{4}=71.4712 \ldots\right)$
attempt to set $2 \pi r$ equal to arc length
$2 \pi r=71.4712 \ldots$

OR
attempt to set $\pi r l$ equal to their area from (a)
$19.5 \pi r=696.844 \ldots$

THEN
$r=11.4\left(=\frac{91}{8}=11.375\right)(\mathrm{cm})$

[Maximum number: 5]

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

(a)

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

[ 2 ]

use of $A=\frac{1}{2} q r \sin \theta$ to obtain $A=\frac{1}{2}(x+2)(5-x)^{2} \sin 30^{\circ}$ $=\frac{1}{4}(x+2)\left(25-10 x+x^{2}\right)$ $A=\frac{1}{4}\left(x^{3}-8 x^{2}+5 x+50\right)$

(b)

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

[ 3 ]

$\quad \mathrm{PQ}=\frac{7}{3}$ (cm) and $\mathrm{PR}=\left(\frac{14}{3}\right)^{2}$ (cm)

\mathrm{QR}^{2}=\left(\frac{7}{3}\right)^{2}+\left(\frac{14}{3}\right)^{4}-2\left(\frac{7}{3}\right)\left(\frac{14}{3}\right)^{2} \cos 30^{\circ}

$=391.702 \ldots$
QR $=19.8(\mathrm{~cm})$

[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 3.1 Geometry and trigonometry - SL content question 3 — IB Maths AA HL

(a)

Find the distance from Bogotá to Moscow.

[ 2 ]

attempt to use Pythagoras' theorem
$\sqrt{12500^{2}+6000^{2}}$ $=13900(\mathrm{~km})(13865.4 \ldots)$

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

[ 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 $064^{\circ}(64.3589 \ldots)$

(c)

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

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.

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

[ 2 ]

(d)

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 $s=r \theta$ $\frac{\theta}{360} \times 2 \pi \times 6=6$ OR $\quad 6=6 \theta \Rightarrow \theta=1$ 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
AG

All question bank results loaded

Covered syllabus points

3.1 Geometry and trigonometry - SL content question 1

Question (a)

Question (b)

3.1 Geometry and trigonometry - SL content question 2

Question (a)

Question (b)

3.1 Geometry and trigonometry - SL content question 3

Question (a)

Question (b)

Question (c)

Question (d)