IB Mathematics AA SL 3.1 Geometry and Trigonometry Question Bank

[Maximum number: 8]

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).

(a)

Find AV .

[ 2 ]

attempt to use the distance formula to find AV

\begin{aligned} & \sqrt{(1-(-1))^{2}+(5-1)^{2}+(0-6)^{2}} \\ & =7.48331 \ldots \\ & =7.48(\mathrm{~cm})(=\sqrt{56} \text { or } 2 \sqrt{14}) \end{aligned}

(b)

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
$=2 \times 7.48 \ldots \times \sin \left(20^{\circ}\right)$ $=5.11888 \ldots$ $=5.12(\mathrm{~cm})$

(c)

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.

(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.

[Maximum number: 16]

The following diagram shows a square ABCD , and a sector OAB of a circle centre O , radius r. Part of the square is shaded and labelled R.

\mathrm{A} \hat{\mathrm{OB}}=\theta, \text { where } 0.5 \leq \theta<\pi

(a)

Show that the area of the square ABCD is $2 r^{2}(1-\cos \theta)$ .

[ 4 ]

area of $A B C D=A B^{2}$ (seen anywhere)
choose cosine rule to find a side of the square
eg $a^{2}=b^{2}+c^{2}-2 b c \cos \theta$
correct substitution (for triangle AOB)
eg $r^{2}+r^{2}-2 \times r \times r \cos \theta, \mathrm{OA}^{2}+\mathrm{OB}^{2}-2 \times \mathrm{OA} \times \mathrm{OB} \cos \theta$
correct working for $\mathrm{AB}^{2}$
eg $2 r^{2}-2 r^{2} \cos \theta$
area $=2 r^{2}(1-\cos \theta)$

NO
Note: Award no marks if the only working is $2 r^{2}-2 r^{2} \cos \theta$ .

(b)

When $\theta=\alpha$ , the area of the square ABCD is equal to the area of the sector OAB .

[ 4 ]

(i) $\frac{1}{2} \alpha r^{2}\left(\right.$ accept $\left.2 r^{2}(1-\cos \alpha)\right)$

(i)

Write down the area of the sector when $\theta=\alpha$ .

$\frac{1}{2} \alpha r^{2}\left(\right.$ accept $\left.2 r^{2}(1-\cos \alpha)\right)$

(ii)

Hence find $\alpha$ .

[ 4 ]

correct equation in one variable
eg $2(1-\cos\alpha)=\frac{1}{2}\alpha$ $\alpha=0.511024$ $\alpha=0.511$ (accept $\theta=0.511$ )

2(1-\cos\alpha)=\frac{1}{2}\alpha,\quad \alpha=0.511

Note: Award A1 for $\alpha=0.511$ and additional answers.

(c)

When $\theta=\beta$ , the area of R is more than twice the area of the sector. Find all possible values of $\beta$ .

[ 8 ]

Note: In this part, accept $\theta$ instead of $\beta$ , and the use of equations instead of inequalities in the working.
attempt to find R
eg subtraction of areas, square - segment
correct expression for segment area
eg $\frac{1}{2} \beta r^{2}-\frac{1}{2} r^{2} \sin \beta$
correct expression for R
eg $\quad 2 r^{2}(1-\cos \beta)-\left(\frac{1}{2} \beta r^{2}-\frac{1}{2} r^{2} \sin \beta\right)$
correct inequality
eg $\quad 2 r^{2}(1-\cos \beta)-\left(\frac{1}{2} \beta r^{2}-\frac{1}{2} r^{2} \sin \beta\right)>2\left(\frac{1}{2} \beta r^{2}\right)$
correct inequality in terms of angle only
eg $\quad 2(1-\cos \beta)-\left(\frac{1}{2} \beta-\frac{1}{2} \sin \beta\right)>\beta$
attempt to solve their inequality, must represent R> twice sector eg sketch, one correct value

Note: Do not award the second (M1) unless the first (M1) for attempting to find R has been awarded.
both correct values 1.30573 and 2.67369
correct inequality $1.31<\beta<2.67$

[Maximum number: 3]

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).

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.

(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
$=2 \times 7.48 \ldots \times \sin \left(20^{\circ}\right)$ $=5.11888 \ldots$ $=5.12(\mathrm{~cm})$

[Maximum number: 3]

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.

Find the radius of the base of the cone.

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}

The following diagram shows a square ABCD , and a sector OAB of a circle centre O , radius r. Part of the square is shaded and labelled R.

\mathrm{A} \hat{\mathrm{OB}}=\theta, \text { where } 0.5 \leq \theta<\pi

When $\theta=\alpha$ , the area of the square ABCD is equal to the area of the sector OAB .

Write down the area of the sector when $\theta=\alpha$ .

$\frac{1}{2} \alpha r^{2}\left(\right.$ accept $\left.2 r^{2}(1-\cos \alpha)\right)$

[Maximum number: 5]

The following diagram shows a straight shoreline, with a supply store at S , a town at T , and an island L .

A boat delivers supplies to the island. The boat leaves S , and sails to the island. Its path makes an angle of $20^{\circ}$ with the shoreline.

It is decided to change the position of the supply store, so that its distance from L is 5 km . The following diagram shows the two possible locations P and Q for the supply store.

Find the size of $S \hat{P} L$ and of $S \hat{Q} L$ .

evidence of choosing sine rule
eg $\quad \frac{\sin A}{a}=\frac{\sin B}{b}, \sin \theta=\frac{(\mathrm{SL}) \sin 20^{\circ}}{5}$
correct substitution
eg $\quad \frac{\sin \theta}{9}=\frac{\sin 20^{\circ}}{5}$
37.9981
$\mathrm{S} \hat{\mathrm{PL}}=38.0^{\circ}$
recognition that second angle is the supplement of first
eg 180-x
142.001
$\mathrm{SQL}=142^{\circ}$

[Maximum number: 3]

Let $f(x)=\cos 2 x$ and $g(x)=2 x^{2}-1$ .

Given that $(g \circ f)(x)$ can be written as $\cos (k x)$ , find the value of $k, k \in \mathbb{Z}$ .

$\quad(g \circ f)(x)=2(\cos (2 x))^{2}-1\left(=2 \cos ^{2}(2 x)-1\right)$
evidence of $2 \cos ^{2} \theta-1=\cos 2 \theta$ (seen anywhere)
$(g \circ f)(x)=\cos 4 x$
k=4

[Maximum number: 6]

Consider the function $f(x)=\frac{2}{\pi} \sin (3 \pi x)+2$ , where $0 \leq x \leq 2$ . The following diagram shows the graph of f.

(a)

Write down the amplitude of f.

[ 1 ]

$0.636619 \ldots$

\text { (amplitude }=\text { ) } 0.637\left(=\frac{2}{\pi}\right)

(b)

Find the period of f.

[ 2 ]

$\frac{2 \pi}{3 \pi}$ OR 3 complete waves every 2 radians

(\text { period }=) 0.667\left(=\frac{2}{3}\right)

(c)

The point P has coordinates (1.63,2.16). State whether P lies above, below or on the graph of f. Justify your answer.

The line $L_{1}$ has equation x-6 y+11=0.

[ 3 ]

substituting to find f(1.63)

(f(1.63)=) 2.21564 \ldots

$2.21564 \ldots>2.16 \Rightarrow$ lies below (the graph of f )

IB Maths AA SL 3.1 Geometry and Trigonometry Sl Content Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question (a)

Question (b)

Question (c)