IB Mathematics AA HL 3.2 Geometry and Trigonometry Question Bank

Practise IB Mathematics AA HL 3.2 by solving advanced vectors, lines, planes, 3D geometry, identities and trigonometric proof problems.

Exam points

HL syllabus concepts and methods
HL extended reasoning

Covered syllabus points

AHL 3.9 (HL)—Reciprocal and inverse trigonometric functions AHL 3.10 (HL)—Compound-angle identities AHL 3.11 (HL)—Further trigonometric equations AHL 3.12 (HL)—Vectors AHL 3.13 (HL)—Scalar product AHL 3.14 (HL)—Vector equations of lines AHL 3.15 (HL)—Line relationships in 3D AHL 3.16 (HL)—Vector product AHL 3.17 (HL)—Planes AHL 3.18 (HL)—Intersections and angles in 3D

[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 3.2 Geometry and trigonometry - AHL content question 1 — 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: 3]

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

(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
$\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=\sin ^{2} 105^{\circ}+2 \sin 105^{\circ} \cos 105^{\circ}+\cos ^{2} 105^{\circ}$ $\left(\sin 105^{\circ}+\cos 105^{\circ}\right)^{2}=1+\sin 210^{\circ}$ $=\frac{1}{2}$ $\sin 105^{\circ}+\cos 105^{\circ}=\frac{1}{\sqrt{2}}$

[Maximum number: 6]

Hence or otherwise solve the equation $\sin x+\sin 3 x=\cos x$ in the interval $0<x<\pi$ .

EITHER
$\sin x+\sin 3 x=\cos x \Rightarrow \frac{1-\cos 4 x}{2 \sin x}=\cos x$ $\Rightarrow 1-\cos 4 x=2 \sin x \cos x,(\sin x \neq 0)$ $\Rightarrow 1-\left(1-2 \sin ^{2} 2 x\right)=\sin 2 x$ $\Rightarrow \sin 2 x(2 \sin 2 x-1)=0$ $\Rightarrow \sin 2 x=0$ or $\sin 2 x=\frac{1}{2}$ $2 x=\pi, 2 x=\frac{\pi}{6}$ and $2 x=\frac{5 \pi}{6}$

OR
$\sin x+\sin 3 x=\cos x \Rightarrow 2 \sin 2 x \cos x=\cos x$ $\Rightarrow(2 \sin 2 x-1) \cos x=0,(\sin x \neq 0)$ $\Rightarrow \sin 2 x=\frac{1}{2}$ or $\cos x=0$ $2 x=\frac{\pi}{6}, 2 x=\frac{5 \pi}{6}$ and $x=\frac{\pi}{2}$

THEN
$\therefore x=\frac{\pi}{2}, x=\frac{\pi}{12}$ and $x=\frac{5 \pi}{12}$

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Covered syllabus points

3.2 Geometry and trigonometry - AHL content question 1

3.2 Geometry and trigonometry - AHL content question 2

3.2 Geometry and trigonometry - AHL content question 3