IB Maths AA HL 3 Geometry and Trigonometry

Practise IB Maths AA HL geometry and trigonometry through shared-core and HL identities, vectors, radians, equations and proof with exact reasoning.

Covered syllabus points

3.1 Geometry and trigonometry - SL content 3.2 Geometry and trigonometry - AHL content

[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 Geometry and trigonometry 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: 1]

Two boats A and B travel due north.
Initially, boat B is positioned 50 metres due east of boat A .
The distances travelled by boat A and boat B , after t seconds, are x metres and y metres respectively. The angle $\theta$ is the radian measure of the bearing of boat B from boat A . This information is shown on the following diagram.

Figure for Question 3 Geometry and trigonometry question 2 — IB Maths AA HL

Show that $y=x+50 \cot \theta$ .

At time T, the following conditions are true.
Boat B has travelled 10 metres further than boat A .
Boat B is travelling at double the speed of boat A .
The rate of change of the angle $\theta$ is -0.1 radians per second.

$\tan \theta=\frac{50}{y-x}$ OR $\cot \theta=\frac{y-x}{50}$ $y=x+50 \cot \theta$
AG
Note: y-x may be identified as a length on a diagram, and not written explicitly.

[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: 8]

(a)

Solve $2 \sin \left(x+60^{\circ}\right)=\cos \left(x+30^{\circ}\right), 0^{\circ} \leq x \leq 180^{\circ}$ .

[ 5 ]

$\quad 2 \sin \left(x+60^{\circ}\right)=\cos \left(x+30^{\circ}\right)$ $2\left(\sin x \cos 60^{\circ}+\cos x \sin 60^{\circ}\right)=\cos x \cos 30^{\circ}-\sin x \sin 30^{\circ}$
(M1)(A1)
$2 \sin x \times \frac{1}{2}+2 \cos x \times \frac{\sqrt{3}}{2}=\cos x \times \frac{\sqrt{3}}{2}-\sin x \times \frac{1}{2}$ $\Rightarrow \frac{3}{2} \sin x=-\frac{\sqrt{3}}{2} \cos x$ $\Rightarrow \tan x=-\frac{1}{\sqrt{3}}$ $\Rightarrow x=150^{\circ}$

(b)

Show that $\sin 105^{\circ}+\cos 105^{\circ}=\frac{1}{\sqrt{2}}$ .

[ 3 ]

(b) EITHER
choosing two appropriate angles, for example $60^{\circ}$ and $45^{\circ}$ $\sin 105^{\circ}=\sin 60^{\circ} \cos 45^{\circ}+\cos 60^{\circ} \sin 45^{\circ}$ and
$\cos 105^{\circ}=\cos 60^{\circ} \cos 45^{\circ}-\sin 60^{\circ} \sin 45^{\circ}$ $\sin 105^{\circ}+\cos 105^{\circ}=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}$ $=\frac{1}{\sqrt{2}}$
OR
attempt to square the expression
$\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=\sin ^{2} 105^{\circ}+2 \sin 105^{\circ} \cos 105^{\circ}+\cos ^{2} 105^{\circ}$ $\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=1+\sin 210^{\circ}$ $=\frac{1}{2}$ $\sin 105^{\circ}+\cos 105^{\circ}=\frac{1}{\sqrt{2}}$

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Covered syllabus points

3 Geometry and trigonometry question 1

Question (a)

Question (b)

3 Geometry and trigonometry question 2

3 Geometry and trigonometry question 3

Question (a)

Question (b)

3 Geometry and trigonometry question 4

Question (a)

Question (b)