G Geometry — printable coverage pack

G1 · Knowledge of perimeter, area, surface area and volume of standard shapes; angle properties of parallel/intersecting lines, triangles, quadrilaterals and polygons; and understand and use circle theorems

Know the standard formulae for perimeters, areas, surface areas and volumes, and keep all measurements in compatible units before substituting.
Angle facts include angles on a line and at a point, vertically opposite and alternate angles, triangle and quadrilateral sums, and interior or exterior angles of polygons.
Circle theorems include: the angle at the centre is twice the angle at the circumference, angles in the same segment, the angle in a semicircle, cyclic quadrilaterals, tangent-radius perpendicularity and the alternate segment theorem.
State the theorem or angle fact when reasons are requested; an unlabelled numerical angle may not earn a reasoning mark.
For a composite shape, split or subtract familiar regions and avoid counting a shared face in a surface-area calculation.

Tier 1 · Easy

1. Work out the interior angle of a regular $12$ -sided polygon.[2 marks]
Answer
- $150^\circ$
Method: The exterior angle is $360^\circ/12=30^\circ$ . An interior angle and its exterior angle sum to $180^\circ$ , so the interior angle is $150^\circ$ .

Tier 2 · Standard

1. A closed cylinder has radius $3$ cm and height $8$ cm. Work out its total surface area in terms of $\pi$ .[3 marks]
Answer
- $66\pi\text{ cm}^2$
Method: The two circular ends have area $2\pi r^2=2\pi(3^2)=18\pi$ . The curved surface has area $2\pi rh=2\pi(3)(8)=48\pi$ . The total is $18\pi+48\pi=66\pi\text{ cm}^2$ .

Tier 3 · Hard

1. Two radii of a circle of radius $10$ cm enclose an angle of $120^\circ$ . Work out the exact area of the minor sector with the triangle formed by the two radii removed.[4 marks]
Answer
- $\left(\frac{100\pi}{3}-25\sqrt{3}\right)\text{ cm}^2$
Method: The sector area is $\frac{120}{360}\pi(10^2)=\frac{100\pi}{3}$ . The triangle area is $\frac12(10)(10)\sin120^\circ=50\cdot\frac{\sqrt3}{2}=25\sqrt3$ . Subtracting gives $\frac{100\pi}{3}-25\sqrt3\text{ cm}^2$ .

G2 · Understand and construct geometrical proofs using formal arguments

A geometrical proof is a connected chain of statements, each supported by a stated theorem, definition or established result.
Use precise labels such as $\angle ABC$ so it is clear which angle each statement concerns.
Congruence can establish equal corresponding sides or angles; name a valid test such as SSS, SAS, ASA or RHS.
Do not assume a diagram is drawn to scale, and do not use the result to be proved as one of your reasons.
For circle proofs, identify the relevant chord, arc, radius or tangent when naming a theorem.

Tier 1 · Easy

1. In isosceles triangle $ABC$ , $AB=AC$ . Point $D$ is the midpoint of $BC$ . Prove that $AD$ is perpendicular to $BC$ .[2 marks]
Answer
- $AD\perp BC$
Method: In triangles $ABD$ and $ACD$ , $AB=AC$ , $BD=DC$ and $AD$ is common. The triangles are congruent by SSS, so $\angle BDA=\angle ADC$ . These equal adjacent angles lie on the straight line $BC$ , so each is $90^\circ$ . Therefore $AD\perp BC$ .

Tier 2 · Standard

1. In a circle with centre $O$ , the minor angle $AOB$ is $2t$ , where $0^\circ<t<90^\circ$ . A tangent is drawn at $A$ . Prove that the acute angle between the tangent and $AB$ is $t$ .[3 marks]
Answer
- The acute angle between the tangent and $AB$ is $t$ .
Method: $OA=OB$ , so triangle $AOB$ is isosceles. Its two base angles are $\frac{180^\circ-2t}{2}=90^\circ-t$ . The tangent is perpendicular to $OA$ , so the angle between the tangent and $AB$ is $90^\circ-(90^\circ-t)=t$ .

Tier 3 · Hard

1. From a point $P$ outside a circle with centre $O$ , tangents touch the circle at $A$ and $B$ . Prove that $OP$ is the perpendicular bisector of $AB$ .[4 marks]
Answer
- $OP$ bisects $AB$ at right angles.
Method: Radii are perpendicular to tangents, so $\angle OAP=\angle OBP=90^\circ$ . Also $OA=OB$ and $OP$ is common, so right triangles $OAP$ and $OBP$ are congruent by RHS. Hence $PA=PB$ . Both $O$ and $P$ are equidistant from $A$ and $B$ , so they lie on the perpendicular bisector of $AB$ . Therefore the line $OP$ is that perpendicular bisector.

G3 · Sine and cosine rules in scalene triangles; area of a triangle = 1/2 ab sinC

Use the sine rule when you know an opposite side-angle pair; keep each side matched with its opposite angle.
Use the cosine rule for three known sides, or for two sides and their included angle.
The area formula $A=\frac12 ab\sin C$ uses the angle $C$ included between sides $a$ and $b$ .
The sine rule can produce two angles because $\sin\theta=\sin(180^\circ-\theta)$ ; check the diagram, interval and triangle angle sum.
Keep full calculator values during working and round only the final answer to the accuracy requested.

Tier 1 · Easy

1. Two sides of a triangle are $8$ cm and $5$ cm, and their included angle is $30^\circ$ . Work out the area.[2 marks]
Answer
- $10\text{ cm}^2$
Method: $A=\frac12 ab\sin C=\frac12(8)(5)\sin30^\circ=20\cdot\frac12=10\text{ cm}^2$ .

Tier 2 · Standard

1. Two sides of a triangle are $7$ cm and $9$ cm, and their included angle is $60^\circ$ . Work out the exact length of the third side.[3 marks]
Answer
- $\sqrt{67}\text{ cm}$
Method: By the cosine rule, $c^2=7^2+9^2-2(7)(9)\cos60^\circ=49+81-63=67$ . A length is positive, so $c=\sqrt{67}\text{ cm}$ .

Tier 3 · Hard

1. In triangle $ABC$ , $A=30^\circ$ , $a=6$ cm and $b=10$ cm. Work out both possible values of angle $C$ . Give each answer to $1$ decimal place.[4 marks]
Answer
- $C=93.6^\circ$ or $C=26.4^\circ$
Method: The sine rule gives $\frac{\sin B}{10}=\frac{\sin30^\circ}{6}$ , so $\sin B=\frac56$ . Thus $B=56.4^\circ$ or $123.6^\circ$ . Using $A+B+C=180^\circ$ gives $C=180^\circ-30^\circ-56.4^\circ=93.6^\circ$ , or $C=180^\circ-30^\circ-123.6^\circ=26.4^\circ$ .

G4 · Use of Pythagoras' theorem in 2D and 3D; recognise Pythagorean triples

In a right-angled triangle, $a^2+b^2=c^2$ , where $c$ is the hypotenuse opposite the right angle.
Common triples such as $3,4,5$ , $5,12,13$ and their multiples can give exact lengths without a calculator.
For a cuboid, find a face diagonal first and then use a second right-angled triangle, or use $d^2=l^2+w^2+h^2$ directly.
Do not use Pythagoras unless a right angle is known, and check that the longest side has been used as the hypotenuse.

Tier 1 · Easy

1. A right-angled triangle has perpendicular sides $5$ cm and $12$ cm. Work out the hypotenuse.[1 mark]
Answer
- $13$ cm
Method: $c=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$ cm. This is the $5,12,13$ triple.

Tier 2 · Standard

1. A cuboid has side lengths $6$ cm, $6$ cm and $7$ cm. Work out the length of its body diagonal.[3 marks]
Answer
- $11$ cm
Method: The body diagonal satisfies $d^2=6^2+6^2+7^2=36+36+49=121$ . Therefore $d=\sqrt{121}=11$ cm.

Tier 3 · Hard

1. A cuboid has base dimensions $8$ cm by $9$ cm and body diagonal $17$ cm. Work out its volume.[4 marks]
Answer
- $864\text{ cm}^3$
Method: If the height is $h$ , three-dimensional Pythagoras gives $8^2+9^2+h^2=17^2$ . Hence $h^2=289-64-81=144$ , so $h=12$ cm. The volume is $8\times9\times12=864\text{ cm}^3$ .

G5 · Apply trigonometry and Pythagoras' theorem to 2 and 3 dimensional problems, including the angle between a line and a plane and between two planes

Draw or identify a right-angled cross-section containing the required line and its perpendicular projection onto the plane.
The angle between a line and a plane is the angle between the line and its projection on that plane, not usually an angle with an arbitrary edge.
For the angle between two planes, use a cross-section perpendicular to their line of intersection; the angle between the two cross-section lines is the required dihedral angle.
Use Pythagoras to find missing face or body diagonals before applying $\sin$ , $\cos$ or $\tan$ .
Mark the right angle and label every calculated length so that the final trigonometric ratio uses the correct triangle.

Tier 1 · Easy

1. A straight ramp is $10$ m long and rises vertically by $6$ m. Work out the angle the ramp makes with the horizontal, to $1$ decimal place.[2 marks]
Answer
- $36.9^\circ$
Method: If the angle is $\theta$ , the rise is opposite and the ramp is the hypotenuse. Thus $\sin\theta=6/10=0.6$ , so $\theta=\sin^{-1}(0.6)=36.9^\circ$ .

Tier 2 · Standard

1. A square-based pyramid has base side $10$ cm. Its apex is vertically above the centre of the base, and a sloping edge from the apex to a base vertex is $13$ cm. Work out the angle that this edge makes with the base, to $1$ decimal place.[3 marks]
Answer
- $57.0^\circ$
Method: The distance from the base centre to a vertex is half the diagonal: $\sqrt{5^2+5^2}=5\sqrt2$ cm. This is the projection of the $13$ cm edge on the base. Hence $\cos\theta=\frac{5\sqrt2}{13}$ , so $\theta=57.0^\circ$ to $1$ decimal place.

Tier 3 · Hard

1. A cuboid $ABCDEFGH$ has $AB=12$ cm, $BC=5$ cm and vertical edge $BF=8$ cm. The base is $ABCD$ and $EFGH$ is the top face. Work out the acute angle between plane $ABGH$ and the base plane $ABCD$ , to $1$ decimal place.[4 marks]
Answer
- $58.0^\circ$
Method: The planes meet along $AB$ . In the cross-section perpendicular to $AB$ , $BC$ lies in the base and $BG$ lies in plane $ABGH$ . Triangle $BCG$ is right-angled with $BC=5$ and $CG=8$ . Therefore $\tan\theta=8/5$ , so $\theta=\tan^{-1}(8/5)=58.0^\circ$ .

G6 · Sketch and use graphs of y = sin x, y = cos x and y = tan x for angles of any size

$y=\sin x$ and $y=\cos x$ have period $360^\circ$ and range $-1\leq y\leq1$ .
$y=\tan x$ has period $180^\circ$ , zeros at multiples of $180^\circ$ , and vertical asymptotes at $90^\circ+180^\circ n$ .
A trigonometric sketch should be smooth and pass through the exact key points; do not join plotted points with straight line segments.
Extend graphs to angles of any size by repeating the correct period in both directions.
Use the graph to read signs, repeated values, intersections and approximate solutions, while excluding points where $\tan x$ is undefined.

Tier 1 · Easy

1. Write down the coordinates of the minimum point of $y=\cos x$ for $0^\circ\leq x\leq360^\circ$ .[1 mark]
Answer
- $(180^\circ,-1)$
Method: The cosine graph starts at $1$ , reaches its minimum value $-1$ halfway through its $360^\circ$ period, then returns to $1$ . The minimum is therefore $(180^\circ,-1)$ .

Tier 2 · Standard

1. Sketch $y=\tan x$ for $0^\circ\leq x\leq360^\circ$ . Mark its zeros and vertical asymptotes.[3 marks]
Answer
- Zeros at $x=0^\circ,180^\circ,360^\circ$ ; vertical asymptotes at $x=90^\circ,270^\circ$ ; increasing tangent branches between them.
Method: Mark zeros every $180^\circ$ and asymptotes $90^\circ$ after each zero. On each interval between consecutive asymptotes, draw a smooth increasing branch from negative to positive values, passing through the relevant zero.

Tier 3 · Hard

1. Use the graphs of $y=\sin x$ and $y=\cos x$ to state the range of $x$ for which both values are positive and $\sin x>\cos x$ , where $0^\circ\leq x\leq360^\circ$ .[3 marks]
Answer
- $45^\circ<x<90^\circ$
Method: Both sine and cosine are positive only in the first quadrant, so $0^\circ<x<90^\circ$ . Their graphs intersect at $x=45^\circ$ . After this intersection and before $90^\circ$ , the sine graph is above the cosine graph. The inequalities are strict, so the endpoints are excluded.

G7 · Use the definitions of sin, cos and tan for any positive angle up to 360 degrees (measured in degrees only)

Use a reference angle and the quadrant to determine the sign: sine is positive in quadrants I and II, cosine in I and IV, and tangent in I and III.
For a point $(x,y)$ at distance $r$ from the origin, $\cos\theta=x/r$ , $\sin\theta=y/r$ and $\tan\theta=y/x$ .
Angles are measured anticlockwise from the positive $x$ -axis and are in degrees for this specification.
When using inverse trigonometric functions, use the quadrant information to convert the calculator's reference angle into the required angle from $0^\circ$ to $360^\circ$ .

Tier 1 · Easy

1. Work out the exact value of $\cos120^\circ$ .[1 mark]
Answer
- $-\frac12$
Method: The reference angle is $60^\circ$ . Cosine is negative in quadrant II, so $\cos120^\circ=-\cos60^\circ=-\frac12$ .

Tier 2 · Standard

1. $180^\circ<\theta<270^\circ$ and $\sin\theta=-\frac{12}{13}$ . Work out the exact values of $\cos\theta$ and $\tan\theta$ .[3 marks]
Answer
- $\cos\theta=-\frac5{13}$
- $\tan\theta=\frac{12}{5}$
Method: A $5,12,13$ triangle gives the remaining magnitude $5/13$ . In quadrant III both sine and cosine are negative, so $\cos\theta=-5/13$ . Then $\tan\theta=\sin\theta/\cos\theta=(-12/13)/(-5/13)=12/5$ .

Tier 3 · Hard

1. $0^\circ<\theta<360^\circ$ , $\tan\theta=-\frac7{24}$ and $\cos\theta>0$ . Work out $\sin\theta$ exactly and $\theta$ to $1$ decimal place.[4 marks]
Answer
- $\sin\theta=-\frac7{25}$
- $\theta=343.7^\circ$
Method: The ratio $7:24$ gives a hypotenuse $25$ . Tangent is negative and cosine positive only in quadrant IV, so sine is negative: $\sin\theta=-7/25$ . The reference angle is $\tan^{-1}(7/24)=16.260^\circ$ . Therefore $\theta=360^\circ-16.260^\circ=343.7^\circ$ to $1$ decimal place.

G8 · Knowledge and use of 30, 60, 90 triangles and 45, 45, 90 triangles

A $30^\circ$ - $60^\circ$ - $90^\circ$ triangle has side ratio $1:\sqrt3:2$ , opposite those angles respectively.
A $45^\circ$ - $45^\circ$ - $90^\circ$ triangle has side ratio $1:1:\sqrt2$ .
These ratios give the exact values $\sin30^\circ=1/2$ , $\cos30^\circ=\sqrt3/2$ and $\sin45^\circ=\cos45^\circ=\sqrt2/2$ .
Keep exact surd values unless a decimal accuracy is requested, and match each side to the angle it is opposite.

Tier 1 · Easy

1. Write down the exact value of $\sin30^\circ$ .[1 mark]
Answer
- $\frac12$
Method: In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the side opposite $30^\circ$ is half the hypotenuse. Therefore $\sin30^\circ=1/2$ .

Tier 2 · Standard

1. A right-angled isosceles triangle has hypotenuse $14$ cm. Work out the exact length of each equal side.[2 marks]
Answer
- $7\sqrt2$ cm
Method: The side ratio is $1:1:\sqrt2$ . If an equal side is $a$ , then $a\sqrt2=14$ , so $a=14/\sqrt2=7\sqrt2$ cm.

Tier 3 · Hard

1. A regular hexagon has side length $8$ cm. Work out the exact distance between a pair of opposite sides and hence the exact area of the hexagon.[4 marks]
Answer
- Distance $=8\sqrt3$ cm
- Area $=96\sqrt3\text{ cm}^2$
Method: Joining the centre to the vertices makes six equilateral triangles. Halving one gives a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle with hypotenuse $8$ and apothem $4\sqrt3$ . The distance between opposite sides is twice the apothem, $8\sqrt3$ . The area is $\frac12\times$ perimeter $\times$ apothem $=\frac12(48)(4\sqrt3)=96\sqrt3\text{ cm}^2$ .

G9 · Know and use tan = sin / cos and sin^2 + cos^2 = 1

Use $\tan x=\frac{\sin x}{\cos x}$ and $\sin^2x+\cos^2x=1$ to replace one trigonometric form with another.
Useful rearrangements include $1-\sin^2x=\cos^2x$ and $1-\cos^2x=\sin^2x$ .
In an identity proof, start from one side and transform it through valid equal expressions; do not assume the target statement.
Factor before substituting identities, and keep brackets around complete numerators and denominators.
Cancellation is valid only where the original expression is defined.

Tier 1 · Easy

1. $x$ is acute and $\sin x=\frac35$ . Work out the exact value of $\tan x$ .[2 marks]
Answer
- $\tan x=\frac34$
Method: $\cos^2x=1-\sin^2x=1-9/25=16/25$ . Since $x$ is acute, $\cos x=4/5$ . Hence $\tan x=(3/5)/(4/5)=3/4$ .

Tier 2 · Standard

1. Simplify $\frac{1-\cos^2x}{\sin x}$ .[2 marks]
Answer
- $\sin x$
Method: Use $1-\cos^2x=\sin^2x$ . Then $\frac{1-\cos^2x}{\sin x}=\frac{\sin^2x}{\sin x}=\sin x$ , for values where the original expression is defined.

Tier 3 · Hard

1. Prove that $\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x}$ for values of $x$ where both sides are defined.[4 marks]
Answer
- $\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x}$
Method: Starting from the left, use the common denominator $\sin x(1+\cos x)$ . The numerator is $\sin^2x+(1+\cos x)^2=\sin^2x+1+2\cos x+\cos^2x=2+2\cos x=2(1+\cos x)$ . Cancelling the common factor gives $2/\sin x$ , as required wherever the original expression is defined.

G10 · Solution of simple trigonometric equations in given intervals

Find a reference angle, use the signs of the trigonometric function to choose the correct quadrants, and list every solution in the stated interval.
For squared equations, rearrange or factor before taking values; do not lose the positive or negative possibilities introduced by a square.
Use the periods $360^\circ$ for sine and cosine and $180^\circ$ for tangent to generate repeated solutions.
Check whether interval endpoints are included and give angles in degrees to the requested accuracy.
Substitute solutions back into the equation when factorisation or cancellation could have introduced or removed values.

Tier 1 · Easy

1. Solve $\sin x=1$ for $0^\circ\leq x\leq360^\circ$ .[1 mark]
Answer
- $x=90^\circ$
Method: The sine graph reaches its maximum value $1$ at $90^\circ$ once in the interval, so $x=90^\circ$ .

Tier 2 · Standard

1. Solve $\cos x=-\frac{\sqrt3}{2}$ for $0^\circ\leq x\leq360^\circ$ .[2 marks]
Answer
- $x=150^\circ$ or $x=210^\circ$
Method: The reference angle is $30^\circ$ . Cosine is negative in quadrants II and III, giving $x=180^\circ-30^\circ=150^\circ$ and $x=180^\circ+30^\circ=210^\circ$ .

Tier 3 · Hard

1. Solve $2\sin^2x+\sin x-1=0$ for $0^\circ\leq x<360^\circ$ .[4 marks]
Answer
- $x=30^\circ,150^\circ,270^\circ$
Method: Let $s=\sin x$ . Then $2s^2+s-1=(2s-1)(s+1)=0$ , so $\sin x=1/2$ or $\sin x=-1$ . In the interval, $\sin x=1/2$ at $30^\circ$ and $150^\circ$ , while $\sin x=-1$ at $270^\circ$ . Therefore the three solutions are $30^\circ,150^\circ,270^\circ$ .