CAIE IGCSE Additional Mathematics Geometry and Trigonometry Question Bank

Practise straight-line and circle geometry, radians, sector measures and trigonometric graphs, identities and equations across coordinate and diagram-based problems.

Covered syllabus points

7. Straight-line graphs 8. Coordinate geometry of the circle 9. Circular measure 10. Trigonometry

[Maximum number: 7]

(a)

A straight line passes through the points (4,23) and (-8,29). Find the point of intersection, P, of this line with the line y=2x+5.

[ 5 ]

$y=-\frac12x+25\text{ isw}$
B3.

M1 for $m=\frac{29-23}{-8-4}$ oe or $-\frac12$ , and M1 FT for $\frac{y-23}{x-4}=$ their $-\frac12$ oe, or y= their $-\frac12x+c$ and $23=-\frac12\times4+c$ oe.

OR M1 for solving 23=4m+c, 29=-8m+c for $m=-\frac12$ or c=25, and M1 FT for correctly using their m or their c to find c or m.

Solves their linear equation simultaneously with y=2x+5 to find x or y. M1: FT their $y=-\frac12x+25$ oe.

(8,21). A1.

(b)

Find the distance of P from the origin.

[ 2 ]

$\sqrt{8^2+21^2}\text{ oe}$
M1: FT their (8,21).

$\sqrt{505}$ isw, or 22.5, or $22.47[22\ldots]$ rot to 2 or more dp. A1.

[Maximum number: 6]

Point A has coordinates (3,-1).
A circle has equation $(x-4)^{2}+(y+3)^{2}=5$ .

(a)

Show that A lies on the circumference of the circle.

[ 1 ]

$(3-4)^{2}+(-1+3)^{2}=5$

or showing that distance between point A and centre of circle = radius e.g.
$\sqrt{(3-4)^{2}+(-1+3)^{2}}=\sqrt{5}$

B1

Accept if x-coordinate substituted to find
y - coordinate

(b)

Given that AB is a diameter of the circle, find the coordinates of B.

[ 2 ]

$\binom{4}{-3}+\binom{1}{-2} \text { oe, soi }$

or using centre and point A e.g.
$\frac{3+x}{2}=4 \text { and } \frac{-1+y}{2}=-3$

or solve simultaneously the line y=-2 x+5 with the circle leading to 3 term quadratic $x^{2}-8 x+15[=0]$ with an attempt to solve

M1

Award M1 for the 2 equations

Award the M1 for the quadratic

(5,-5)

A1

(c)

Find the equation of the tangent to the circle at A.

[ 3 ]

gradient of radius $=\frac{-3+1}{4-3}=-2$ soi by gradient
of tangent

M1

gradient of tangent $=\frac{-1}{\text { their }(-2)}$

M1

FT their -2
Allow if differentiation is used e.g.:

$y+1=\frac{1}{2}(x-3)$ oe

A1

$\text { FT } \frac{-1}{\text { their }-2}$

ISW from a correct unsimplified answer

Alternative

use of differentiation e.g.:
$\begin{aligned} (y+3)^{2}=5-(x-4)^{2} y=\sqrt{-x^{2}+8 x-11}-3 \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-2 x+8}{2 \sqrt{-x^{2}+8 x-11}} \end{aligned}$

(M1)

allow one error
or use of implicit differentiation (not on syllabus) e.g.:
$2 x+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}-8+6 \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
leading to $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{8-2 x}{2 y+6}$

Substitute (3,-1) in their $\frac{\mathrm{d} y}{\mathrm{~d} x}$
to get gradient $=\frac{1}{2}$

(M1)

Dep on first M1

$y+1=\frac{1}{2}(x-3)$ oe

(A1)

ISW from a correct unsimplified answer

[Maximum number: 3]

A circle, centre O and radius $r\text{ cm}$ , has a sector OAB of fixed area $10\text{ cm}^2$ . Angle AOB is $\theta$ radians and the perimeter of the sector is $P\text{ cm}$ .

Find an expression for P in terms of r.

$\frac12r^2\theta=10,\quad \theta=\frac{20}{r^2}$
B1
$P=2r+r\theta$
M1
$P=2r+\frac{20}{r}$
A1

[Maximum number: 7]

(a)

Figure for Question (a) — CAIE IGCSE Additional Math

The diagram shows an equilateral triangle ABC with side a.
M is the midpoint of AC and angle $AMB=90^\circ$ .
Use the diagram to find $\sec 30^\circ$ .

[ 3 ]

$\frac{2}{\sqrt{3}}$ or $\frac{2\sqrt{3}}{3}$
B2 for $[BM=]\frac{\sqrt{3}a}{2}$ or $\sec^2 30=1+\left(\frac{1}{\sqrt{3}}\right)^2$ or $\cos^2 30=1-\left(\frac{a/2}{a}\right)^2$ oe
B1 for $[BM^2=]a^2-\frac{a^2}{4}$ soi

(b)

Show that $\frac{1}{\sec x-1}+\frac{1}{\sec x+1}$ can be written as $2\operatorname{cosec}x\cot x$ .

[ 4 ]

$\frac{2\sec x}{(\sec x-1)(\sec x+1)}$ soi
B1 for adding the two fractions.
$\frac{2\sec x}{\sec^2 x-1}$
B1 dep on previous B1 for expanding and simplifying the denominator.
$\frac{2\sec x}{\tan^2 x}$ or $\frac{2}{\tan^2 x\cos x}$
B1 dep on previous B1 for use of $\sec^2 x=1+\tan^2 x$ .
$\frac{2\sec x}{\tan x}\times\cot x$ or $\frac{2\cos x}{\sin^2 x}$ oe or $\frac{2}{\tan x\sin x}$ , leading to $2\operatorname{cosec}x\cot x$ AG
B1 dep on previous B1 with at least one more correct step to get to the given answer.
Alternative method:
$\begin{aligned} \frac{1}{\frac{1}{\cos x}-1}+\frac{1}{\frac{1}{\cos x}+1}&=\frac{\cos x}{1-\cos x}+\frac{\cos x}{1+\cos x} \end{aligned}$
(B1)
$=\frac{\cos x+\cos^2 x+\cos x-\cos^2 x}{1-\cos^2 x}$
(B1) dep on previous B1 for adding the two fractions.
$=\frac{2\cos x}{\sin^2 x}$
(B1) dep on previous B1 for use of $\sin^2 x=1-\cos^2 x$ .
$=2\operatorname{cosec}x\times\frac{\cos x}{\sin x}=2\operatorname{cosec}x\cot x$ AG
(B1)

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Covered syllabus points

Geometry and Trigonometry question 1

Question (a)

Question (b)

Geometry and Trigonometry question 2

Question (a)

Question (b)

Question (c)

Geometry and Trigonometry question 3

Geometry and Trigonometry question 4

Question (a)

Question (b)