IB Maths AA SL 3.1 Geometry and Trigonometry Question Bank

Practise modelling three-dimensional shapes, selecting triangle rules, working in radians, simplifying identities and solving trigonometric models over stated intervals.

Exam points

combine volume, surface-area and Pythagoras formulae to model composite solids and unknown dimensions
select sine rule, cosine rule or area formulae for triangles, bearings and ambiguous-case solutions
use radian arc-length and sector-area formulae to solve for angles, radii, perimeters or areas
derive exact trigonometric values and simplify identities using unit-circle and double-angle relationships
fit or interpret sinusoidal graphs, then solve trigonometric equations over every stated interval

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]

All lengths in this question are in centimetres.
A solid metal ornament is in the shape of a right pyramid, with vertex V and square base ABCD . The centre of the base is X . Point V has coordinates (1,5,0) and point A has coordinates (-1,1,6).

Figure for Question 3.1 Geometry and trigonometry - SL content question 1 — IB Maths AA SL

(a)

Given that $\mathrm{A} \hat{\mathrm{V}} \mathrm{B}=40^{\circ}$ , find AB .

The volume of the pyramid is $57.2 \mathrm{~cm}^{3}$ , correct to three significant figures.

[ 3 ]

(b) METHOD 1
attempt to apply cosine rule OR sine rule to find AB
$(\mathrm{AB}=) \sqrt{7.48 \ldots^{2}+7.48 \ldots^{2}-2 \times 7.48 \ldots \times 7.48 \ldots \cos \left(40^{\circ}\right)}$ OR $\frac{\mathrm{AB}}{\sin 40^{\circ}}=\frac{\sqrt{56}}{\sin 70^{\circ}}$ $=5.11888 \ldots$ $=5.12(\mathrm{~cm})$
METHOD 2
Let M be the midpoint of $[\mathrm{AB}]$
attempt to apply right-angled trigonometry on triangle AVM
$=2 \times 7.48 \ldots \times \sin \left(20^{\circ}\right)$ $=5.11888 \ldots$ $=5.12(\mathrm{~cm})$

(b)

Find the height of the pyramid, VX.

A second ornament is in the shape of a cuboid with a rectangular base of length $2 x \mathrm{~cm}$ , width $x \mathrm{~cm}$ and height $y \mathrm{~cm}$ . The cuboid has the same volume as the pyramid.

[ 3 ]

METHOD 1
equating volume of pyramid formula to 57.2
$\frac{1}{3} \times 5.11 \ldots{ }^{2} \times h=57.2$
(A1)
$h=6.54886 \ldots$ $h=6.55(\mathrm{~cm})$
METHOD 2
Let M be the midpoint of AB ]
$A V^{2}=A M^{2}+M X^{2}+X V^{2}$ $\Rightarrow X V=\sqrt{7.48 \ldots^{2}-\left(\frac{5.11 \ldots}{2}\right)^{2}-\left(\frac{5.11 \ldots}{2}\right)^{2}}$ $h=6.54886 \ldots$ $h=6.55(\mathrm{~cm})$

[Maximum number: 5]

A monument is in the shape of a right cone with a vertical height of 20 metres. Oliver stands 5 metres from the base of the monument. His eye level is 1.8 metres above the ground and the angle of elevation from Oliver's eye level to the vertex of the cone is $58^{\circ}$ , as shown on the following diagram.

Figure for Question 3.1 Geometry and trigonometry - SL content question 2 — IB Maths AA SL

(a)

Find the radius of the base of the cone.

[ 3 ]

attempt to use trigonometry to find the radius of the cone OR Oliver's distance from
centre (r+5)

\begin{aligned} & \tan 58^{\circ}=\frac{18.2}{r+5} \text { OR } \frac{r+5}{\sin 32^{\circ}}=\frac{18.2}{\sin 58^{\circ}} \text { OR }(r+5=) 11.3726 \ldots \\ & r=6.37262 \ldots(\mathrm{~m}) \\ & (r=) 6.37(\mathrm{~m}) \end{aligned}

(b)

Find the volume of the monument.

[ 2 ]

attempt to substitute h=20 and their radius into the correct volume of cone formula

\begin{aligned} & V=\frac{\pi(6.37262 \ldots)^{2}(20)}{3} \\ & =850.540 \ldots \\ & =851\left(\mathrm{~m}^{3}\right) \end{aligned}

Note: Accept $849.840 \ldots(850)$ obtained from using r=6.37.

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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)