G Geometry — revision question pack

10 specification points · notes, questions, answers and worked methods

Checked against AQA 8365 section G. Review basis: the qualification registry sourced from the AQA Level 2 Certificate in Further Mathematics (8365) specification; registry verification recorded 11 July 2026.

How this checking works

Answer all questions in the spaces provided.

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

Explanation

Standard areas include rectangle $lw$ , triangle $\frac12bh$ , parallelogram $bh$ , trapezium $\frac12(a+b)h$ and circle $\pi r^2$ ; circumference is $2\pi r$ . Prism volume is cross-sectional area times length; cylinder volume is $\pi r^2h$ , curved area $2\pi rh$ and total area $2\pi rh+2\pi r^2$ .
Sphere volume and area are $\frac43\pi r^3$ and $4\pi r^2$ ; cone volume and total area are $\frac13\pi r^2h$ and $\pi rl+\pi r^2$ ; pyramid volume is $\frac13\times\text{base area}\times\text{perpendicular height}$ . Prism and pyramid surface areas sum all exposed faces.
Angle properties cover parallel and intersecting lines, triangles, all special quadrilaterals and polygons.
Circle facts include centre angle twice circumference angle, same-segment angles, semicircle right angle, opposite cyclic angles, radius–tangent perpendicularity, alternate segment, centre-perpendicular bisecting a chord, and equal tangents from an external point.
Examiners expect compatible units and named reasons.

An angle at the circumference is subtended by a chord joining its two endpoints.

Worked example

Opposite angles of a cyclic quadrilateral are $112^\circ$ and $x^\circ$ . Work out $x$ and give a reason.

1.All four vertices lie on one circle, so the quadrilateral is cyclic.
2.Opposite angles in a cyclic quadrilateral sum to $180^\circ$ .
3. $x=180^\circ-112^\circ=68^\circ$ .

Answer: $x=68^\circ$ , because opposite angles in a cyclic quadrilateral sum to $180^\circ$ .

Common mistakes

Don't use $360^\circ-112^\circ$ for one opposite angle instead of the cyclic total $180^\circ$ .
Don't give the correct angle but omit the circle theorem when a reason is required.
Don't include a shared internal face when calculating the external surface area of a composite solid.

Exam tip

For a reasoning angle question, write the named theorem on the same line as the equation it justifies.

Tier 1 · Easy

1
Work out the interior angle of a regular $12$ -sided polygon.
[2 marks]
2
An angle at the centre of a circle is $146^\circ$ . Work out the angle at the circumference subtended by the same arc.
[1 mark]

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]
2
A solid is made by placing a hemisphere of radius $4$ cm on a cylinder of radius $4$ cm. The total volume is $\frac{560\pi}{3}\text{ cm}^3$ . Work out the height of the cylinder.
[3 marks]
3
The interior angle of a regular polygon is five times its exterior angle. Work out the number of sides of the polygon.
[3 marks]

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]
2
Points $A$ , $B$ , $C$ and $D$ occur in that order on a circle. The angle between the tangent at $A$ and chord $AB$ , measured on the side of $AB$ that does not contain $C$ and $D$ , is $38^\circ$ , and $\angle BAC=27^\circ$ . Work out $\angle ABC$ , giving geometrical reasons.
[4 marks]
3
A cone of radius $3$ cm and slant height $5$ cm is joined to the top of a cylinder of radius $3$ cm and height $h$ cm. The circular join is not exposed. The total external surface area is $60\pi\text{ cm}^2$ . Work out the total volume of the solid.
[4 marks]
4
A cylinder has radius $3$ cm and height $12$ cm. A sphere has the same volume as the cylinder. Work out the radius of the sphere, giving an exact answer.
[3 marks]
5
From a point $P$ outside a circle with centre $O$ , tangents $PA$ and $PB$ touch the circle. The smaller angle $APB$ is $46^\circ$ . Point $C$ lies on the major arc $AB$ . Work out $\angle ACB$ , giving geometrical reasons.
[3 marks]

G2 · Understand and construct geometrical proofs using formal arguments

Explanation

A geometrical proof is a connected chain of statements, each justified by a definition, theorem or established result. Precise labels such as $\angle ABC$ identify the vertex and arms unambiguously.
Congruence can prove corresponding sides or angles equal; valid tests include SSS, SAS, ASA and RHS. Similarity, parallel-line facts and circle theorems may also supply steps.
A diagram must not be assumed to scale, and the desired conclusion cannot be used as a premise.
For circle proofs, the relevant chord, radius, tangent or arc should be identified.
Examiners expect reasons beside key statements and a final sentence that explicitly establishes the proposition.

A diagonal divides parallelogram ABCD into two triangles for a congruence proof.

Worked example

In parallelogram $ABCD$ , diagonal $AC$ is drawn. Prove that triangles $ABC$ and $CDA$ are congruent.

1. $AB=CD$ and $BC=AD$ because opposite sides of a parallelogram are equal.
2. $AC$ is common to both triangles.
3.The three corresponding side pairs are equal, so the triangles are congruent by SSS.

Answer: $\triangle ABC\cong\triangle CDA$ by SSS.

Common mistakes

Don't claim the triangles are congruent because they look equal on the diagram.
Don't name SSS without identifying all three corresponding side pairs.
Don't list triangle vertices in an order that mismatches corresponding points.

Exam tip

For congruence, state each matching side or angle pair before naming the test.

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]
2
$ABCD$ is a rhombus. Prove that diagonal $AC$ bisects $\angle DAB$ .
[2 marks]

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]
2
The diagonals $AC$ and $BD$ of parallelogram $ABCD$ intersect at $E$ . Prove that $AE=EC$ .
[3 marks]
3
In triangle $ABC$ , $AB=AC$ and $AD$ bisects $\angle BAC$ , where $D$ lies on $BC$ . Prove that $D$ is the midpoint of $BC$ .
[3 marks]

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]
2
In quadrilateral $ABCD$ , $AB\parallel CD$ and $AB=CD$ . Prove that $ABCD$ is a parallelogram.
[4 marks]
3
Points $P$ , $Q$ , $R$ and $S$ lie in that order on a circle. The chords $PR$ and $QS$ intersect at $X$ . Prove that the angles of triangle $PXQ$ are equal to the angles of triangle $SXR$ .
[4 marks]
4
In quadrilateral $ABCD$ , $AB=AD$ and $CB=CD$ . Diagonals $AC$ and $BD$ intersect at $X$ . Prove that $AC$ is perpendicular to $BD$ .
[4 marks]
5
A circle has centre $O$ , and $A$ and $B$ are two points on the circle. The tangents at $A$ and at $B$ meet at the external point $T$ . Prove that $TA=TB$ .
[4 marks]

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

Explanation

Right-angled triangles use $\sin=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$ and $\tan=\frac{\text{opposite}}{\text{adjacent}}$ ; inverse ratios recover an unknown acute angle from side lengths. For a scalene triangle, the sine rule $\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}$ needs an opposite side–angle pair.
The cosine rule $c^2=a^2+b^2-2ab\cos C$ uses three sides or two sides and their included angle.
Area is $\frac12ab\sin C$ .
A sine-rule angle may have an alternative because $\sin\theta=\sin(180^\circ-\theta)$ ; the triangle sum decides validity.
Examiners expect correctly paired labels, unrounded working and final rounding only as requested.

In a scalene triangle, each lowercase side is opposite its matching uppercase angle.

Worked example

Two sides of a triangle are $7$ cm and $10$ cm and their included angle is $48^\circ$ . Work out the third side to $3$ significant figures.

1.Use the cosine rule: $c^2=7^2+10^2-2(7)(10)\cos48^\circ$ .
2.Evaluate $c^2\approx55.3217$ .
3.Take the positive square root and round only the final value.

Answer: $c=7.44\text{ cm}$ to $3$ significant figures.

Common mistakes

Don't use the sine rule when no opposite side–angle pair is known.
Don't pair the $48^\circ$ angle with a non-included pair of sides in the cosine rule.
Don't round $\cos48^\circ$ early and change the final length.

Exam tip

Mark the included angle and its opposite side on the diagram before selecting the cosine-rule form.

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]
2
In triangle $ABC$ , $A=30^\circ$ , $B=45^\circ$ and the side opposite $A$ is $6$ cm. Work out the exact length of the side opposite $B$ .
[2 marks]

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]
2
Two sides of a triangle are $8$ cm and $12$ cm. Its area is $24\sqrt3\text{ cm}^2$ , and the included angle $C$ is acute. Work out $C$ .
[3 marks]
3
In triangle $ABC$ , $A=42^\circ$ , $B=68^\circ$ and the side opposite $A$ is $9$ cm. Work out the length of the side opposite $C$ , giving the answer to $3$ significant figures.
[3 marks]

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]
2
In triangle $ABC$ , $AB=10$ cm, $AC=14$ cm and the area is $35\sqrt3\text{ cm}^2$ . Angle $A$ is obtuse. Work out the exact length of $BC$ .
[4 marks]
3
A triangle has side lengths $5$ cm, $7$ cm and $8$ cm. Work out its exact area.
[4 marks]
4
In quadrilateral $ABCD$ , vertices $B$ and $D$ lie on opposite sides of diagonal $AC$ . The lengths are $AB=7$ cm, $BC=9$ cm, $CD=8$ cm, $DA=6$ cm and $AC=10$ cm. Work out the exact area of the quadrilateral.
[4 marks]
5
Two sides of a triangle are $x$ cm and $(x+1)$ cm, where $x>0$ . The angle between them is $60^\circ$ and the opposite side has length $\sqrt7$ cm. Work out the exact area of the triangle.
[4 marks]

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

Explanation

In a right-angled triangle, $a^2+b^2=c^2$ , where $c$ is the hypotenuse opposite the right angle. The theorem can find a missing side or verify a right angle.
Required familiar triples include $3,4,5$ , $5,12,13$ , $8,15,17$ and $7,24,25$ , together with their multiples.
In three dimensions, a face diagonal can be found first and used in a second right-angled triangle; for a cuboid this combines to $d^2=l^2+w^2+h^2$ .
Pythagoras is valid only when the right angle is established, and a calculated length must be positive.
Examiners expect the longest side to be treated as the hypotenuse and each 3D right-angled cross-section to be identified.

A cuboid’s body diagonal satisfies d² = l² + w² + h².

Worked example

A cuboid has side lengths $3$ cm, $4$ cm and $12$ cm. Work out its body diagonal.

1.Apply three-dimensional Pythagoras: $d^2=3^2+4^2+12^2$ .
2. $d^2=9+16+144=169$ .
3.Take the positive square root.

Answer: $d=13\text{ cm}$ .

Common mistakes

Don't add the three side lengths instead of their squares.
Don't use only two dimensions and report the face diagonal $5$ cm.
Don't use Pythagoras on a triangle without identifying a right angle.

Exam tip

In 3D, mark the face projection and body diagonal before writing the Pythagorean relation.

Tier 1 · Easy

1
A right-angled triangle has perpendicular sides $5$ cm and $12$ cm. Work out the hypotenuse.
[1 mark]
2
Show that a triangle with side lengths $9$ cm, $40$ cm and $41$ cm is right-angled.
[2 marks]

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]
2
A vertical pole is $12$ m high and stands at one corner of a rectangular field measuring $9$ m by $8$ m. Work out the distance from the top of the pole to the opposite corner of the field.
[3 marks]
3
The points $A=(1,-2)$ and $B=(k,10)$ are $13$ units apart. Given that $k>1$ , work out $k$ .
[3 marks]

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]
2
A right triangular prism is $84$ cm long. Its triangular ends have side lengths $5$ cm, $12$ cm and $13$ cm. Work out the longest distance between two vertices of the prism.
[4 marks]
3
A right-angled triangle has side lengths $(x+1)$ cm, $(x+8)$ cm and $(x+9)$ cm, where $x>0$ . Work out its area.
[4 marks]
4
The convex trapezium $ABCD$ has $AB\parallel CD$ , $\angle DAB=\angle ADC=90^\circ$ , $AB=25$ cm, $CD=10$ cm and $BC=17$ cm. Work out its area.
[3 marks]
5
A square-based pyramid has base side $10$ cm. Its apex is vertically above the centre of the base and is $13$ cm from each base vertex. Work out the exact volume of the pyramid.
[4 marks]

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

Explanation

A 3D trigonometry problem should be reduced to a labelled right-angled cross-section.
The angle between a line and a plane is the angle between the line and its perpendicular projection onto that plane, not an angle with an arbitrary edge.
For the angle between two planes, take a cross-section perpendicular to their line of intersection; the angle between the two cross-section lines is the dihedral angle.
Pythagoras often supplies a face diagonal or projection before sine, cosine or tangent is applied.
Examiners expect the relevant projection, right angle and required angle to be identified, because using a visually convenient but incorrect triangle gives the wrong ratio.

The angle between a line and a plane is measured against the line’s perpendicular projection in the plane.

Worked example

A cuboid has base dimensions $6$ cm by $8$ cm and height $5$ cm. Work out the angle between its body diagonal and the base, to $1$ decimal place.

1.The body diagonal projects onto the base diagonal, whose length is $\sqrt{6^2+8^2}=10$ cm.
2.In the right-angled cross-section, $\tan\theta=\frac{5}{10}=0.5$ .
3. $\theta=\tan^{-1}(0.5)$ .

Answer: $\theta=26.6^\circ$ to $1$ decimal place.

Common mistakes

Don't measure the body diagonal’s angle against a $6$ cm base edge instead of its $10$ cm projection.
Don't use the body diagonal as the adjacent side in the tangent ratio.
Don't find the complementary angle with the vertical height.

Exam tip

For a line–plane angle, identify the line’s projection on the plane before choosing a trigonometric ratio.

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]
2
A symmetrical roof is $10$ m wide and its ridge is $2$ m above the horizontal ceiling. Work out the angle between either roof face and the ceiling, to $1$ decimal place.
[2 marks]

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]
2
The base of a cuboid measures $3$ cm by $4$ cm and its vertical height is $12$ cm. Work out the acute angle between its body diagonal and a vertical edge, to $1$ decimal place.
[3 marks]
3
A vertical mast stands at the centre of a rectangular level courtyard measuring $16$ m by $30$ m. A straight wire joins the top of the $8$ m mast to a corner of the courtyard. Work out the angle between the wire and the courtyard, to $1$ decimal place.
[3 marks]

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]
2
A square-based pyramid has base side $12$ cm and its apex is $8$ cm vertically above the centre of the base. Work out the slant height of a triangular face and the angle between that face and the base, to $1$ decimal place.
[4 marks]
3
The base $ABCD$ of a box is a rectangle, with the vertices in order, $AB=16$ cm and $BC=12$ cm. Point $P$ is $9$ cm vertically above $A$ , and $M$ is the midpoint of $CD$ . Work out the angle between $PM$ and the base, to $1$ decimal place.
[4 marks]
4
A vertical mast $OT$ is $12$ m high. Points $A$ and $B$ are on horizontal ground, $OA=9$ m, $OB=16$ m and $\angle AOB=90^\circ$ . Work out the angle between the wires $TA$ and $TB$ to $1$ decimal place.
[4 marks]
5
Cuboid $ABCDEFGH$ has rectangular base $ABCD$ , with $E$ , $F$ , $G$ and $H$ directly above $A$ , $B$ , $C$ and $D$ respectively. Given $AB=8$ cm, $BC=6$ cm and $BF=12$ cm, work out the acute angle between plane $ACF$ and the base plane $ABCD$ to $1$ decimal place.
[4 marks]

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

Explanation

In degrees, $y=\sin x$ and $y=\cos x$ have period $360^\circ$ and range $-1\leq y\leq1$ .
Sine passes through $0$ at multiples of $180^\circ$ ; cosine begins at $1$ when $x=0^\circ$ .
The tangent graph has period $180^\circ$ , zeros at multiples of $180^\circ$ and vertical asymptotes at $90^\circ+180^\circ n$ .
Graphs extend to angles of any size by repeating these periods in both directions.
Examiners expect smooth curves through exact key points, separate tangent branches that never cross their asymptotes, and correct use of graphs to read signs, intersections or approximate solutions.

Sine and cosine repeat every 360°; tangent repeats every 180° between vertical asymptotes.

Worked example

State the zeros and maximum point of $y=\sin x$ for $0^\circ\leq x\leq360^\circ$ .

1.Sine is zero at the start, halfway point and end of one period.
2.Its maximum value $1$ occurs one quarter of the way through the period.
3.Attach the corresponding angle coordinates.

Answer: Zeros at $(0^\circ,0)$ , $(180^\circ,0)$ and $(360^\circ,0)$ ; maximum at $(90^\circ,1)$ .

Common mistakes

Don't use period $180^\circ$ for sine or cosine.
Don't draw tangent continuously through a vertical asymptote.
Don't join trigonometric key points with straight line segments.

Exam tip

Mark zeros, extrema and asymptotes first, then draw smooth periodic branches through them.

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]
2
Use the graph of $y=\sin x$ to write down the value of $\sin450^\circ$ .
[1 mark]

Tier 2 · Standard

1
Sketch $y=\tan x$ for $0^\circ\leq x\leq360^\circ$ . Mark its zeros and vertical asymptotes.
[3 marks]
2
Use the graph of $y=\sin x$ to solve $\sin x=\frac12$ for $-180^\circ\leq x\leq540^\circ$ .
[3 marks]
3
Use the graphs of $y=\sin x$ and $y=\cos x$ to work out the $x$ -coordinates of all their intersections for $-360^\circ\leq x\leq720^\circ$ .
[3 marks]

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]
2
Use the graphs of $y=\sin x$ and $y=\tan x$ to state all ranges of $x$ for which $\sin x>0$ and $\tan x<0$ , where $-360^\circ\leq x\leq360^\circ$ .
[3 marks]
3
Use the graph of $y=\cos x$ to work out the coordinates of every maximum, minimum and $x$ -axis crossing for $-180^\circ\leq x\leq540^\circ$ .
[4 marks]
4
Use the graph of $y=\sin x$ . Solve $\sin x=-0.6$ in the interval $-360^\circ\leq x\leq180^\circ$ . Give each value to $1$ decimal place.
[3 marks]
5
Use the graph of $y=\tan x$ . Work out every range satisfying $\tan x\geq1$ in the interval $-360^\circ\leq x\leq360^\circ$ .
[4 marks]

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

Explanation

For any positive angle up to $360^\circ$ , measured anticlockwise from the positive $x$ -axis, a point $(x,y)$ at distance $r$ from the origin gives $\cos\theta=\frac xr$ , $\sin\theta=\frac yr$ and $\tan\theta=\frac yx$ .
Sine is positive in quadrants I and II, cosine in I and IV, and tangent in I and III.
A reference angle gives the magnitude, while the quadrant fixes the sign and full angle.
Inverse trigonometric functions often return only a principal value, so quadrant information must be applied.
Examiners expect degree mode and an angle within the stated quadrant or interval.

Coordinate definitions and sign patterns for sine, cosine and tangent in four quadrants.

Worked example

The terminal point of angle $\theta$ is $(-5,12)$ , with $0^\circ<\theta<180^\circ$ . Work out $\sin\theta$ , $\cos\theta$ and $\tan\theta$ exactly.

1. $r=\sqrt{(-5)^2+12^2}=13$ .
2. $\sin\theta=\frac{y}{r}=\frac{12}{13}$ and $\cos\theta=\frac{x}{r}=-\frac5{13}$ .
3. $\tan\theta=\frac{y}{x}=\frac{12}{-5}=-\frac{12}{5}$ .

Answer: $\sin\theta=\frac{12}{13}$ , $\cos\theta=-\frac5{13}$ and $\tan\theta=-\frac{12}{5}$ .

Common mistakes

Don't use distance $r=17$ by adding the coordinate magnitudes instead of applying Pythagoras.
Don't make cosine positive even though the point lies in quadrant II.
Don't measure the angle clockwise from the positive $x$ -axis.

Exam tip

Write the quadrant sign and calculate $r$ before forming the three coordinate ratios.

Tier 1 · Easy

1
Work out the exact value of $\cos120^\circ$ .
[1 mark]
2
$OP$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin and $P=(-3,4)$ . Work out the exact value of $\sin\theta$ .
[2 marks]

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]
2
$OP$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin. The point $P$ has $x$ -coordinate $-8$ and is $17$ units from $O$ . Given that $90^\circ<\theta<180^\circ$ , work out the $y$ -coordinate of $P$ and the value of $\theta$ to $1$ decimal place.
[3 marks]
3
$OP$ has length $12$ units and makes an angle of $330^\circ$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin. Work out the exact coordinates of $P$ .
[3 marks]

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]
2
$OP$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin and $P=(k+4,k-2)$ . The angle $\theta$ lies between $270^\circ$ and $360^\circ$ , and $\tan\theta=-\frac12$ . Work out $k$ and the exact values of $\sin\theta$ and $\cos\theta$ .
[4 marks]
3
$OP$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin and $90^\circ<\theta<180^\circ$ . Point $P=(x,y)$ lies on both $x^2+y^2=100$ and $x+3y=10$ . Work out the coordinates of $P$ and the exact values of $\sin\theta$ , $\cos\theta$ and $\tan\theta$ .
[4 marks]
4
$OP$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin and $P=(x,y)$ . Given $x:y=-5:12$ , $OP=39$ units and $90^\circ<\theta<180^\circ$ , work out the coordinates of $P$ and the value of $\theta$ to $1$ decimal place.
[3 marks]
5
$OP$ makes an angle $\theta$ with the positive $x$ -axis, measured anticlockwise, where $O$ is the origin, $P=(x,y)$ and $315^\circ<\theta<360^\circ$ . Point $Q=(x,0)$ , $OP=13$ units and triangle $OPQ$ has area $30$ square units. Work out the coordinates of $P$ and the exact values of $\sin\theta$ , $\cos\theta$ and $\tan\theta$ .
[4 marks]

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

Explanation

A $30^\circ$ – $60^\circ$ – $90^\circ$ triangle has side ratio $1:\sqrt3:2$ , opposite $30^\circ$ , $60^\circ$ and $90^\circ$ respectively. A $45^\circ$ – $45^\circ$ – $90^\circ$ triangle has ratio $1:1:\sqrt2$ .
These ratios follow by halving an equilateral triangle or bisecting a square, and they generate exact trigonometric values such as $\sin30^\circ=\frac12$ , $\cos30^\circ=\frac{\sqrt3}{2}$ and $\sin45^\circ=\frac{\sqrt2}{2}$ .
They also produce exact lengths in regular polygons and composite geometry.
Scale every side by the same factor and retain exact surds unless a decimal is requested.
Examiners expect each ratio length to be matched to the angle opposite it.

The side ratios are 1:√3:2 for a 30°–60°–90° triangle and 1:1:√2 for a 45°–45°–90° triangle.

Worked example

A $30^\circ$ – $60^\circ$ – $90^\circ$ triangle has hypotenuse $10$ cm. Work out the other two side lengths exactly.

1.The side ratio is $1:\sqrt3:2$ .
2.The scale factor is $10\div2=5$ .
3.Multiply the two leg ratios by $5$ .

Answer: The shorter leg is $5$ cm and the longer leg is $5\sqrt3$ cm.

Common mistakes

Don't place the side of ratio $1$ opposite $60^\circ$ instead of $30^\circ$ .
Don't scale the hypotenuse to $10$ but leave the other ratio lengths unscaled.
Don't round $5\sqrt3$ when an exact length is requested.

Exam tip

Write the ratio beside the three corresponding sides before applying the common scale factor.

Tier 1 · Easy

1
Write down the exact value of $\sin30^\circ$ .
[1 mark]
2
A $45^\circ$ - $45^\circ$ - $90^\circ$ triangle has one shorter side of length $6$ cm. Work out its hypotenuse exactly.
[1 mark]

Tier 2 · Standard

1
A right-angled isosceles triangle has hypotenuse $14$ cm. Work out the exact length of each equal side.
[2 marks]
2
An equilateral triangle has side length $12$ cm. Work out its height and area exactly.
[3 marks]
3
A prism has length $12$ cm. Its cross-section is a right-angled isosceles triangle with hypotenuse $10$ cm. Work out the volume of the prism.
[3 marks]

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]
2
A square $ABCD$ has side length $6$ cm. An equilateral triangle $ABE$ is constructed outside the square. Work out the exact length of $EC$ .
[4 marks]
3
A regular octagon has side length $4$ cm. Work out its exact area.
[4 marks]
4
The axial cross-section of a cone is an equilateral triangle of side length $12$ cm. Work out the exact volume of the cone.
[4 marks]
5
The convex trapezium $ABCD$ has $AB\parallel CD$ , with $AB=20$ cm, $CD<AB$ and perpendicular height $6$ cm. The interior angles satisfy $\angle DAB=60^\circ$ and $\angle ABC=45^\circ$ . Work out the exact area of the trapezium.
[4 marks]

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

Explanation

The identities $\tan x=\frac{\sin x}{\cos x}$ and $\sin^2x+\cos^2x=1$ allow one trigonometric form to be replaced by another. Useful rearrangements are $1-\sin^2x=\cos^2x$ and $1-\cos^2x=\sin^2x$ .
They also solve equations by replacing squared terms or tangent with sine and cosine before factorising. When finding a missing value, the quadrant determines the sign of a square root.
In an identity proof, start from one side and transform it through valid equal expressions until the other is reached.
A common denominator often exposes an identity.
Examiners expect restrictions to be respected whenever a cancelled denominator could be zero.

Worked example

$180^\circ<x<270^\circ$ and $\cos x=-\frac{12}{13}$ . Work out $\sin x$ and $\tan x$ exactly.

1. $\sin^2x=1-\cos^2x=1-\frac{144}{169}=\frac{25}{169}$ .
2.Quadrant III makes sine negative, so $\sin x=-\frac5{13}$ .
3. $\tan x=\frac{\sin x}{\cos x}=\frac{-5/13}{-12/13}=\frac5{12}$ .

Answer: $\sin x=-\frac5{13}$ and $\tan x=\frac5{12}$ .

Common mistakes

Don't take only the positive square root for sine despite the quadrant III restriction.
Don't write $\tan x=\frac{\cos x}{\sin x}$ and obtain the reciprocal.
Don't cancel a trigonometric factor without noting values where the original denominator is zero.

Exam tip

After using $\sin^2x+\cos^2x=1$ , apply the quadrant sign before taking the final square root.

Tier 1 · Easy

1
$x$ is acute and $\sin x=\frac35$ . Work out the exact value of $\tan x$ .
[2 marks]
2
Simplify $\sin^2x+\cos^2x+\frac{\tan x\cos x}{\sin x}$ for values of $x$ where the expression is defined.
[2 marks]

Tier 2 · Standard

1
Simplify $\frac{1-\cos^2x}{\sin x}$ .
[2 marks]
2
Given that $\sin x+\cos x=\frac12$ , work out the exact value of $\sin x\cos x$ .
[3 marks]
3
Given that $\tan x=2$ , work out the exact values of $\sin^2x$ and $\cos^2x$ .
[3 marks]

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]
2
Prove that $(\sin x+\cos x)^4+(\sin x-\cos x)^4=2+8\sin^2x\cos^2x$ .
[4 marks]
3
Prove that $\frac{\tan x+\sin x}{\tan x-\sin x}=\left(\frac{1+\cos x}{\sin x}\right)^2$ for values of $x$ where both sides are defined.
[4 marks]
4
Given that $90^\circ<x<135^\circ$ and $\sin x-\cos x=\frac75$ , work out the exact values of $\sin x$ , $\cos x$ and $\tan x$ .
[4 marks]
5
The angle $x$ is acute and $\tan x+\frac1{\tan x}=\frac{10}{3}$ . Work out $\sin x+\cos x$ exactly.
[3 marks]

G10 · Solution of simple trigonometric equations in given intervals

Explanation

To solve a simple trigonometric equation, first isolate the trigonometric function and find a reference angle. Use its sign to choose the correct quadrants, then list every solution in the stated interval.
Sine and cosine repeat every $360^\circ$ ; tangent repeats every $180^\circ$ .
A squared equation may produce positive and negative function values, so factor or take both square-root possibilities.
Interval endpoints must be included or excluded exactly as stated, and angles are measured in degrees for this specification.
Examiners expect all solutions, normally in ascending order, with substitution checks when factorisation or cancellation could introduce or remove a value.

The equation cos x = 1/2 has solutions in quadrants I and IV over one full turn.

Worked example

Solve $2\cos x=1$ for $0^\circ\leq x<360^\circ$ .

1.Isolate cosine: $\cos x=\frac12$ .
2.The reference angle is $60^\circ$ .
3.Cosine is positive in quadrants I and IV, giving $60^\circ$ and $360^\circ-60^\circ$ .

Answer: $x=60^\circ$ or $x=300^\circ$ .

Common mistakes

Don't report only the principal calculator value $60^\circ$ .
Don't choose quadrant II because sine, not cosine, is positive there.
Don't include $360^\circ$ when the interval endpoint is strict.

Exam tip

Write the reference angle and permitted quadrants before listing every solution in the interval.

Tier 1 · Easy

1
Solve $\sin x=1$ for $0^\circ\leq x\leq360^\circ$ .
[1 mark]
2
Solve $\tan x=0$ for $0^\circ\leq x<360^\circ$ .
[2 marks]

Tier 2 · Standard

1
Solve $\cos x=-\frac{\sqrt3}{2}$ for $0^\circ\leq x\leq360^\circ$ .
[2 marks]
2
Solve $\tan x=\sqrt3$ for $0^\circ\leq x<360^\circ$ .
[3 marks]
3
Solve $\sin x=-\frac{\sqrt2}{2}$ for $0^\circ\leq x<360^\circ$ .
[3 marks]

Tier 3 · Hard

1
Solve $2\sin^2x=3\cos x$ for $0^\circ\leq x<360^\circ$ .
[4 marks]
2
For real $k$ , the equation $\sin x=k$ is considered on the interval $0^\circ\leq x\leq360^\circ$ . Find every value of $k$ for which the equation has exactly one solution, and state the corresponding solution in each case.
[4 marks]
3
Solve $\sin^2x=3\cos^2x$ for $0^\circ\leq x<360^\circ$ .
[4 marks]
4
Solve $\tan x=-\frac43$ for $0^\circ\leq x<360^\circ$ . Give each value to $1$ decimal place.
[3 marks]
5
Solve $\frac{\sin x}{1+\cos x}=1$ in the interval $0^\circ\leq x<360^\circ$ , where the expression is defined.
[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

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

Tier 1 · Easy

Mark scheme for G1 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$150^\circ$	2	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 $150^\circ$ .
2	$73^\circ$	1	The angle at the centre is twice the angle at the circumference subtended by the same arc. Therefore the required angle is $146^\circ\div2=73^\circ$ .

Tier 2 · Standard

Mark scheme for G1 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$66\pi\text{ cm}^2$	3	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 $18\pi+48\pi=66\pi\text{ cm}^2$ .
2	$9$ cm	3	The hemisphere has volume $\frac12\times\frac43\pi(4^3)=\frac{128\pi}{3}\text{ cm}^3$ . The cylinder therefore has volume $\frac{560\pi}{3}-\frac{128\pi}{3}=144\pi\text{ cm}^3$ . If its height is $h$ , then $16\pi h=144\pi$ , so $h=9$ cm.
3	$12$ sides	3	Let the exterior angle be $e^\circ$ . The adjacent interior and exterior angles sum to $180^\circ$ , so $5e+e=180$ and $e=30$ . The number of sides is $360\div30=12$ .

Tier 3 · Hard

Mark scheme for G1 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\left(\frac{100\pi}{3}-25\sqrt{3}\right)\text{ cm}^2$	4	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 $\frac{100\pi}{3}-25\sqrt3\text{ cm}^2$ .
2	$\angle ABC=115^\circ$ , using the alternate segment theorem ( $\angle ADB=38^\circ$ ), angles in the same segment ( $\angle BDC=27^\circ$ ) and opposite angles of a cyclic quadrilateral summing to $180^\circ$	4	By the alternate segment theorem, $\angle ADB=38^\circ$ . Angles subtended by chord $BC$ in the same segment are equal, so $\angle BDC=\angle BAC=27^\circ$ . Hence $\angle ADC=38^\circ+27^\circ=65^\circ$ . Opposite angles of cyclic quadrilateral $ABCD$ sum to $180^\circ$ , so $\angle ABC=180^\circ-65^\circ=115^\circ$ .
3	$66\pi\text{ cm}^3$	4	The external area is the curved cylinder area, the cylinder's lower circular end and the curved cone area: $6\pi h+9\pi+15\pi=60\pi$ . Hence $h=6$ . The cone's perpendicular height is $\sqrt{5^2-3^2}=4$ cm. The total volume is $\pi(3^2)(6)+\frac13\pi(3^2)(4)=54\pi+12\pi=66\pi\text{ cm}^3$ .
4	$3\sqrt[3]{3}$ cm	3	The cylinder volume is $\pi(3^2)(12)=108\pi\text{ cm}^3$ . If the sphere radius is $r$ cm, then $\frac43\pi r^3=108\pi$ , giving $r^3=81$ . Therefore $r=\sqrt[3]{81}=3\sqrt[3]{3}$ cm.
5	$\angle AOB=134^\circ$ because the radii are perpendicular to the tangents; therefore $\angle ACB=67^\circ$ because the angle at the centre is twice the angle at the circumference	3	The radii $OA$ and $OB$ are perpendicular to the tangents, so $\angle OAP=\angle OBP=90^\circ$ . Angles in quadrilateral $OAPB$ sum to $360^\circ$ , giving $\angle AOB=360^\circ-90^\circ-90^\circ-46^\circ=134^\circ$ . Since $C$ is on the major arc, $\angle ACB$ stands on the minor arc $AB$ and is half the angle at the centre. Hence $\angle ACB=67^\circ$ .

G2 · Understand and construct geometrical proofs using formal arguments

Tier 1 · Easy

Mark scheme for G2 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$AD\perp BC$	2	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 $AD\perp BC$ .
2	$\angle DAC=\angle CAB$	2	In triangles $ADC$ and $ABC$ , $AD=AB$ and $DC=BC$ because all sides of a rhombus are equal, while $AC$ is common. The triangles are congruent by SSS, so $\angle DAC=\angle CAB$ . Therefore $AC$ bisects $\angle DAB$ .

Tier 2 · Standard

Mark scheme for G2 Tier 2 · Standard
Q	Answer	Mark	Comments
1	The acute angle between the tangent and $AB$ is $t$ .	3	$OA=OB$ , 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 $90^\circ-(90^\circ-t)=t$ .
2	$AE=EC$	3	Since $AB\parallel CD$ , $\angle BAE=\angle DCE$ and $\angle ABE=\angle CDE$ by alternate angles. Also $AB=CD$ because opposite sides of a parallelogram are equal. Thus triangles $ABE$ and $CDE$ are congruent by ASA, so corresponding sides $AE$ and $EC$ are equal.
3	$BD=DC$ , so $D$ is the midpoint of $BC$ .	3	In triangles $ABD$ and $ACD$ , $AB=AC$ , $\angle BAD=\angle DAC$ and $AD$ is common. The triangles are congruent by SAS, so the corresponding sides satisfy $BD=DC$ . Since $D$ lies on $BC$ , it is the midpoint of $BC$ .

Tier 3 · Hard

Mark scheme for G2 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$OP$ bisects $AB$ at right angles.	4	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.
2	$ABCD$ is a parallelogram.	4	Draw diagonal $AC$ . Since $AB\parallel CD$ , $\angle BAC=\angle DCA$ by alternate angles. Also $AB=CD$ and $AC$ is common, so triangles $BAC$ and $DCA$ are congruent by SAS. Hence $\angle BCA=\angle DAC$ . These are alternate angles, so $BC\parallel AD$ . Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram.
3	$\angle XPQ=\angle XSR$ and $\angle XQP=\angle XRS$ (angles in the same segment), and $\angle PXQ=\angle SXR$ (vertically opposite), so the triangles have equal angles	4	Chord $QR$ subtends $\angle QPR$ at $P$ and $\angle QSR$ at $S$ , which lie in the same segment, so $\angle XPQ=\angle XSR$ . Chord $PS$ subtends $\angle PQS$ at $Q$ and $\angle PRS$ at $R$ , also in the same segment, so $\angle XQP=\angle XRS$ . Finally $\angle PXQ=\angle SXR$ because vertically opposite angles are equal. Therefore each angle of triangle $PXQ$ equals the corresponding angle of triangle $SXR$ .
4	$\triangle ABC\cong\triangle ADC$ by SSS, so $\angle BAX=\angle XAD$ ; then $\triangle ABX\cong\triangle ADX$ by SAS, so the equal adjacent angles $\angle BXA$ and $\angle AXD$ are each $90^\circ$ and $AC\perp BD$	4	Triangles $ABC$ and $ADC$ have $AB=AD$ , $BC=DC$ and common side $AC$ , so they are congruent by SSS. Hence $\angle BAC=\angle CAD$ . In triangles $ABX$ and $ADX$ , $AB=AD$ , $AX$ is common and $\angle BAX=\angle XAD$ , so they are congruent by SAS. Therefore $\angle BXA=\angle AXD$ . These equal adjacent angles lie on the straight line $BD$ , so each is $90^\circ$ and $AC\perp BD$ .
5	$\angle OAT=\angle OBT=90^\circ$ (a tangent is perpendicular to the radius at the point of contact), $OA=OB$ (radii) and $OT$ is common, so $\triangle OAT\equiv\triangle OBT$ by RHS congruence; corresponding sides give $TA=TB$	4	Join $OA$ , $OB$ and $OT$ . A tangent is perpendicular to the radius at the point of contact, so $\angle OAT=\angle OBT=90^\circ$ . In triangles $OAT$ and $OBT$ : $OA=OB$ because both are radii, and $OT$ is a common side which is the hypotenuse of both right-angled triangles. The triangles are therefore congruent by RHS. Corresponding sides give $TA=TB$ .

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

Tier 1 · Easy

Mark scheme for G3 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$10\text{ cm}^2$	2	$A=\frac12 ab\sin C=\frac12(8)(5)\sin30^\circ=20\cdot\frac12=10\text{ cm}^2$ .
2	$6\sqrt2\text{ cm}$	2	By the sine rule, $\frac{b}{\sin45^\circ}=\frac{6}{\sin30^\circ}$ . Hence $b=6\times\frac{\sqrt2/2}{1/2}=6\sqrt2\text{ cm}$ .

Tier 2 · Standard

Mark scheme for G3 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\sqrt{67}\text{ cm}$	3	By the cosine rule, $c^2=7^2+9^2-2(7)(9)\cos60^\circ=49+81-63=67$ . A length is positive, so $c=\sqrt{67}\text{ cm}$ .
2	$C=60^\circ$	3	Using the area formula, $24\sqrt3=\frac12(8)(12)\sin C=48\sin C$ . Thus $\sin C=\sqrt3/2$ . The acute angle with this sine value is $C=60^\circ$ .
3	$12.6$ cm	3	$C=180^\circ-42^\circ-68^\circ=70^\circ$ . By the sine rule, $\frac{c}{\sin70^\circ}=\frac{9}{\sin42^\circ}$ , so $c=\frac{9\sin70^\circ}{\sin42^\circ}=12.639\ldots$ cm. To $3$ significant figures, $c=12.6$ cm.

Tier 3 · Hard

Mark scheme for G3 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$C=93.6^\circ$ or $C=26.4^\circ$	4	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 $C=180^\circ-30^\circ-123.6^\circ=26.4^\circ$ .
2	$BC=2\sqrt{109}\text{ cm}$	4	$35\sqrt3=\frac12(10)(14)\sin A$ , so $\sin A=\sqrt3/2$ . Since $A$ is obtuse, $A=120^\circ$ . The cosine rule gives $BC^2=10^2+14^2-2(10)(14)\cos120^\circ=296+140=436$ . Therefore $BC=\sqrt{436}=2\sqrt{109}\text{ cm}$ .
3	$10\sqrt3\text{ cm}^2$	4	Let $C$ be the angle between the $5$ cm and $7$ cm sides. The cosine rule gives $8^2=5^2+7^2-2(5)(7)\cos C$ , so $\cos C=1/7$ . Hence $\sin C=\sqrt{1-1/49}=4\sqrt3/7$ . The area is $\frac12(5)(7)\sin C=10\sqrt3\text{ cm}^2$ .
4	$\left(24+6\sqrt{26}\right)\text{ cm}^2$	4	Triangle $ACD$ is right-angled because $6^2+8^2=10^2$ , so its area is $\frac12(6)(8)=24\text{ cm}^2$ . In triangle $ABC$ , the cosine rule gives $\cos BAC=(7^2+10^2-9^2)/(2\cdot7\cdot10)=17/35$ . Hence $\sin BAC=\sqrt{1-(17/35)^2}=6\sqrt{26}/35$ . The area of triangle $ABC$ is $\frac12(7)(10)(6\sqrt{26}/35)=6\sqrt{26}\text{ cm}^2$ . Adding the two triangle areas gives $(24+6\sqrt{26})\text{ cm}^2$ .
5	$\frac{3\sqrt3}{2}\text{ cm}^2$	4	The cosine rule gives $7=x^2+(x+1)^2-2x(x+1)\cos60^\circ=x^2+x+1$ . Hence $x^2+x-6=0$ , so $(x+3)(x-2)=0$ . Since $x>0$ , $x=2$ and the two sides are $2$ cm and $3$ cm. The area is $\frac12(2)(3)\sin60^\circ=\frac{3\sqrt3}{2}\text{ cm}^2$ .

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

Tier 1 · Easy

Mark scheme for G4 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$13$ cm	1	$c=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$ cm. This is the $5,12,13$ triple.
2	It is right-angled because $9^2+40^2=41^2$ .	2	The longest side is $41$ cm. Now $9^2+40^2=81+1600=1681$ , and $41^2=1681$ . The converse of Pythagoras' theorem therefore shows that the triangle is right-angled.

Tier 2 · Standard

Mark scheme for G4 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$11$ cm	3	The body diagonal satisfies $d^2=6^2+6^2+7^2=36+36+49=121$ . Therefore $d=\sqrt{121}=11$ cm.
2	$17$ m	3	The squared ground distance to the opposite corner is $9^2+8^2=145$ . Using the pole as the third perpendicular dimension, the required distance $d$ satisfies $d^2=145+12^2=289$ . Hence $d=17$ m.
3	$k=6$	3	The horizontal and vertical differences are $k-1$ and $12$ . Pythagoras gives $(k-1)^2+12^2=13^2$ , so $(k-1)^2=25$ and $k-1=\pm5$ . The root $k=-4$ is rejected because $k>1$ , so $k-1=5$ , giving $k=6$ .

Tier 3 · Hard

Mark scheme for G4 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$864\text{ cm}^3$	4	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 $8\times9\times12=864\text{ cm}^3$ .
2	$85$ cm	4	The greatest separation between two vertices on one triangular end is $13$ cm. This direction is perpendicular to the prism length, so the longest vertex-to-vertex distance $d$ satisfies $d^2=84^2+13^2=7056+169=7225$ . Therefore $d=85$ cm, since $84^2+13^2=7225=85^2$ .
3	$30\text{ cm}^2$	4	The longest side is $(x+9)$ cm, so $(x+1)^2+(x+8)^2=(x+9)^2$ . Expanding and simplifying gives $x^2=16$ . Since $x>0$ , $x=4$ , so the perpendicular sides are $5$ cm and $12$ cm. The area is $\frac12(5)(12)=30\text{ cm}^2$ .
4	$140\text{ cm}^2$	3	The difference between the parallel side lengths is $25-10=15$ cm. The side $BC$ , this $15$ cm difference and the perpendicular height form a right-angled triangle. The height is $\sqrt{17^2-15^2}=8$ cm. Therefore the trapezium area is $\frac12(25+10)(8)=140\text{ cm}^2$ .
5	$\frac{100\sqrt{119}}{3}\text{ cm}^3$	4	The distance from the centre of the square to a vertex is half its diagonal, $5\sqrt2$ cm. If the perpendicular height is $h$ cm, then $h^2+(5\sqrt2)^2=13^2$ , so $h=\sqrt{119}$ cm. The base area is $10^2=100\text{ cm}^2$ . Therefore the volume is $\frac13(100)(\sqrt{119})=\frac{100\sqrt{119}}3\text{ cm}^3$ .

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

Tier 1 · Easy

Mark scheme for G5 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$36.9^\circ$	2	If the angle is $\theta$ , the rise is opposite and the ramp is the hypotenuse. Thus $\sin\theta=6/10=0.6$ , so $\theta=\sin^{-1}(0.6)=36.9^\circ$ .
2	$21.8^\circ$	2	A cross-section perpendicular to the ridge has horizontal run $10/2=5$ m and vertical rise $2$ m. If the angle is $\theta$ , then $\tan\theta=2/5$ . Hence $\theta=\tan^{-1}(2/5)=21.8^\circ$ .

Tier 2 · Standard

Mark scheme for G5 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$57.0^\circ$	3	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.
2	$22.6^\circ$	3	The body diagonal is $\sqrt{3^2+4^2+12^2}=13$ cm. In the right-angled cross-section, the $12$ cm vertical edge is adjacent to the required angle and the body diagonal is the hypotenuse. Thus $\cos\theta=12/13$ , giving $\theta=22.6^\circ$ .
3	$25.2^\circ$	3	The wire's projection on the courtyard runs from the centre to a corner, so its length is $\sqrt{(16/2)^2+15^2}=\sqrt{8^2+15^2}=17$ m. If the required angle is $\theta$ , then $\tan\theta=8/17$ . Hence $\theta=25.2^\circ$ to $1$ decimal place.

Tier 3 · Hard

Mark scheme for G5 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$58.0^\circ$	4	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 $\theta=\tan^{-1}(8/5)=58.0^\circ$ .
2	Slant height $=10$ cm Angle $=53.1^\circ$	4	Take the cross-section through the apex, the base centre and the midpoint of a base edge. The horizontal distance from the centre to that midpoint is $6$ cm and the vertical height is $8$ cm. The slant height is $\sqrt{6^2+8^2}=10$ cm. The cross-section is perpendicular to the shared base edge, so it shows the angle between the planes. Thus $\tan\theta=8/6$ , giving $\theta=53.1^\circ$ .
3	$32.0^\circ$	4	The projection of $PM$ on the base is $AM$ . From $A$ to $M$ , the components in the base are $8$ cm parallel to $AB$ and $12$ cm parallel to $BC$ , so $AM=\sqrt{8^2+12^2}=4\sqrt{13}$ cm. Therefore $\tan\theta=9/(4\sqrt{13})$ , giving $\theta=32.0^\circ$ to $1$ decimal place.
4	$61.3^\circ$	4	Pythagoras gives $TA=\sqrt{12^2+9^2}=15$ m and $TB=\sqrt{12^2+16^2}=20$ m. On the ground, $AB=\sqrt{9^2+16^2}=\sqrt{337}$ m. In triangle $ATB$ , the cosine rule gives $\cos ATB=(15^2+20^2-337)/(2\cdot15\cdot20)=12/25$ . Hence $\angle ATB=61.3^\circ$ to $1$ decimal place.
5	$68.2^\circ$	4	The planes meet along $AC$ . Let $N$ be the foot of the perpendicular from $B$ to $AC$ . Since triangle $ABC$ has area $24\text{ cm}^2$ and $AC=10$ cm, $\frac12(10)(BN)=24$ , so $BN=4.8$ cm. Both $BN$ and $FN$ are perpendicular to $AC$ , so $\angle FNB$ is the angle between the planes. In right triangle $FBN$ , $\tan\angle FNB=BF/BN=12/4.8=2.5$ , giving $68.2^\circ$ .

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

Tier 1 · Easy

Mark scheme for G6 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$(180^\circ,-1)$	1	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 $(180^\circ,-1)$ .
2	$1$	1	The sine graph has period $360^\circ$ , and $450^\circ=90^\circ+360^\circ$ . Therefore $\sin450^\circ=\sin90^\circ=1$ .

Tier 2 · Standard

Mark scheme for G6 Tier 2 · Standard
Q	Answer	Mark	Comments
1	Zeros at $x=0^\circ,180^\circ,360^\circ$ ; vertical asymptotes at $x=90^\circ,270^\circ$ ; increasing tangent branches between them.	3	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.
2	$x=30^\circ,150^\circ,390^\circ,510^\circ$	3	In one period, $\sin x=1/2$ at $30^\circ$ and $150^\circ$ . Repeating the sine graph $360^\circ$ later gives $390^\circ$ and $510^\circ$ . The corresponding values one period earlier are below $-180^\circ$ , so the four listed values are all the solutions in the interval.
3	$x=-315^\circ,-135^\circ,45^\circ,225^\circ,405^\circ,585^\circ$	3	The graphs first intersect at $x=45^\circ$ , and corresponding intersections repeat every $180^\circ$ . The values $45^\circ+180^\circ n$ in the interval occur for $n=-2,-1,0,1,2,3$ , giving $-315^\circ,-135^\circ,45^\circ,225^\circ,405^\circ,585^\circ$ .

Tier 3 · Hard

Mark scheme for G6 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$45^\circ<x<90^\circ$	3	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.
2	$-270^\circ<x<-180^\circ$ or $90^\circ<x<180^\circ$	3	The sine graph is positive on $-360^\circ<x<-180^\circ$ and $0^\circ<x<180^\circ$ . On those intervals, the tangent graph is negative only on $-270^\circ<x<-180^\circ$ and $90^\circ<x<180^\circ$ . The endpoints are excluded because sine is zero or tangent is undefined there.
3	Maxima: $(0^\circ,1)$ and $(360^\circ,1)$ ; minima: $(-180^\circ,-1)$ , $(180^\circ,-1)$ and $(540^\circ,-1)$ ; crossings: $(-90^\circ,0)$ , $(90^\circ,0)$ , $(270^\circ,0)$ and $(450^\circ,0)$	4	Cosine has period $360^\circ$ . Its maxima occur at multiples of $360^\circ$ , its minima occur $180^\circ$ after a maximum, and its zeros occur at $90^\circ+180^\circ n$ . Restricting these patterns to the stated closed interval gives the two maxima, three minima and four crossings listed.
4	$x=-143.1^\circ$ or $x=-36.9^\circ$	3	The reference angle is $\sin^{-1}(0.6)=36.869\ldots^\circ$ . In one negative cycle the sine graph reaches $-0.6$ at $-180^\circ+36.869\ldots^\circ=-143.130\ldots^\circ$ and at $-36.869\ldots^\circ$ . These are the only crossings in the interval, so $x=-143.1^\circ$ or $-36.9^\circ$ .
5	$-315^\circ\leq x<-270^\circ$ , $-135^\circ\leq x<-90^\circ$ , $45^\circ\leq x<90^\circ$ or $225^\circ\leq x<270^\circ$	4	The tangent graph has period $180^\circ$ and equals $1$ at $45^\circ+180^\circ n$ . On each increasing branch it is at least $1$ from this point up to, but not including, the next vertical asymptote at $90^\circ+180^\circ n$ . Restricting these branches to the interval gives the four stated ranges.

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

Tier 1 · Easy

Mark scheme for G7 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$-\frac12$	1	The reference angle is $60^\circ$ . Cosine is negative in quadrant II, so $\cos120^\circ=-\cos60^\circ=-\frac12$ .
2	$\sin\theta=\frac45$	2	The distance from the origin is $r=\sqrt{(-3)^2+4^2}=5$ . By the coordinate definition, $\sin\theta=y/r=4/5$ .

Tier 2 · Standard

Mark scheme for G7 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\cos\theta=-\frac5{13}$ $\tan\theta=\frac{12}{5}$	3	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 $\tan\theta=\sin\theta/\cos\theta=(-12/13)/(-5/13)=12/5$ .
2	$y=15$ $\theta=118.1^\circ$	3	Pythagoras gives $y^2=17^2-(-8)^2=225$ . In quadrant II, $y$ is positive, so $y=15$ . Also $\cos\theta=x/r=-8/17$ , so $\theta=\cos^{-1}(-8/17)=118.1^\circ$ .
3	$P=(6\sqrt3,-6)$	3	The coordinates are $(12\cos330^\circ,12\sin330^\circ)$ . Since $\cos330^\circ=\sqrt3/2$ and $\sin330^\circ=-1/2$ , this gives $P=(6\sqrt3,-6)$ .

Tier 3 · Hard

Mark scheme for G7 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\sin\theta=-\frac7{25}$ $\theta=343.7^\circ$	4	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.
2	$k=0$ $\sin\theta=-\frac{\sqrt5}{5}$ (or $-\frac{1}{\sqrt5}$ ) $\cos\theta=\frac{2\sqrt5}{5}$ (or $\frac{2}{\sqrt5}$ )	4	$\tan\theta=\frac{k-2}{k+4}=-\frac12$ , so $2k-4=-k-4$ and $k=0$ . Thus $P=(4,-2)$ and $r=\sqrt{4^2+(-2)^2}=2\sqrt5$ . Therefore $\sin\theta=-2/(2\sqrt5)=-\sqrt5/5$ and $\cos\theta=4/(2\sqrt5)=2\sqrt5/5$ . Both values match an angle between $270^\circ$ and $360^\circ$ , where sine is negative and cosine is positive.
3	$P=(-8,6)$ $\sin\theta=\frac35$ , $\cos\theta=-\frac45$ , $\tan\theta=-\frac34$	4	Substitute $x=10-3y$ into $x^2+y^2=100$ : $(10-3y)^2+y^2=100$ , so $10y(y-6)=0$ . The solutions are $(10,0)$ and $(-8,6)$ ; only $(-8,6)$ gives an angle between $90^\circ$ and $180^\circ$ . Since $OP=10$ , $\sin\theta=6/10=3/5$ , $\cos\theta=-8/10=-4/5$ and $\tan\theta=6/(-8)=-3/4$ .
4	$P=(-15,36)$ and $\theta=112.6^\circ$	3	Write $x=-5k$ and $y=12k$ with $k>0$ because $P$ is in quadrant II. Then $OP=\sqrt{25k^2+144k^2}=13k=39$ , so $k=3$ and $P=(-15,36)$ . The reference angle is $\tan^{-1}(36/15)=67.380\ldots^\circ$ , hence $\theta=180^\circ-67.380\ldots^\circ=112.6^\circ$ .
5	$P=(12,-5)$ $\sin\theta=-\frac5{13}$ , $\cos\theta=\frac{12}{13}$ and $\tan\theta=-\frac5{12}$	4	In quadrant IV, let $v=-y>0$ . The triangle area gives $\frac12xv=30$ , so $xv=60$ , while $OP=13$ gives $x^2+v^2=169$ . Hence $(x+v)^2=169+120=289$ , so $x+v=17$ . From $x+v=17$ , substitute $v=17-x$ into $xv=60$ : $x(17-x)=60$ , so $x^2-17x+60=0$ and $(x-5)(x-12)=0$ , giving $x=5$ or $x=12$ . Since $315^\circ<\theta<360^\circ$ , $x>v$ , so $x=12$ , $v=5$ and $P=(12,-5)$ . Dividing the coordinates by $OP=13$ gives the three exact ratios.

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

Tier 1 · Easy

Mark scheme for G8 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\frac12$	1	In a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, the side opposite $30^\circ$ is half the hypotenuse. Therefore $\sin30^\circ=1/2$ .
2	$6\sqrt2$ cm	1	The side ratio is $1:1:\sqrt2$ . Scaling a shorter side from $1$ to $6$ makes the hypotenuse $6\sqrt2$ cm.

Tier 2 · Standard

Mark scheme for G8 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$7\sqrt2$ cm	2	The side ratio is $1:1:\sqrt2$ . If an equal side is $a$ , then $a\sqrt2=14$ , so $a=14/\sqrt2=7\sqrt2$ cm.
2	Height $=6\sqrt3$ cm Area $=36\sqrt3\text{ cm}^2$	3	The altitude bisects the triangle into two $30^\circ$ - $60^\circ$ - $90^\circ$ triangles with hypotenuse $12$ and short side $6$ . The height is therefore $6\sqrt3$ cm. The area is $\frac12(12)(6\sqrt3)=36\sqrt3\text{ cm}^2$ .
3	$300\text{ cm}^3$	3	In a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, each shorter side is $10/\sqrt2=5\sqrt2$ cm. The cross-sectional area is $\frac12(5\sqrt2)^2=25\text{ cm}^2$ . The volume is $25\times12=300\text{ cm}^3$ .

Tier 3 · Hard

Mark scheme for G8 Tier 3 · Hard
Q	Answer	Mark	Comments
1	Distance $=8\sqrt3$ cm Area $=96\sqrt3\text{ cm}^2$	4	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 $=\frac12(48)(4\sqrt3)=96\sqrt3\text{ cm}^2$ .
2	$EC=6\sqrt{2+\sqrt3}\text{ cm}$ (or $3(\sqrt6+\sqrt2)\text{ cm}$ )	4	The altitude of equilateral triangle $ABE$ is $3\sqrt3$ cm, and it meets $AB$ at its midpoint. Relative to $E$ , point $C$ is therefore $3$ cm horizontally away and $6+3\sqrt3$ cm vertically away. Hence $EC^2=3^2+(6+3\sqrt3)^2=72+36\sqrt3=36(2+\sqrt3)$ . Thus $EC=6\sqrt{2+\sqrt3}$ cm, which is also $3(\sqrt6+\sqrt2)$ cm.
3	$32(1+\sqrt2)\text{ cm}^2$ (or $(32+32\sqrt2)\text{ cm}^2$ )	4	Enclose the octagon in a square. Each removed corner is a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle with hypotenuse $4$ cm, so each shorter side is $2\sqrt2$ cm. The enclosing square has side $4+4\sqrt2$ cm and area $(4+4\sqrt2)^2=48+32\sqrt2$ . The four removed triangles have total area $4\times\frac12(2\sqrt2)^2=16$ , leaving $32+32\sqrt2=32(1+\sqrt2)\text{ cm}^2$ .
4	$72\sqrt3\pi\text{ cm}^3$	4	The base of the axial cross-section is the cone's diameter, so the radius is $6$ cm. Halving the equilateral triangle produces a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle with hypotenuse $12$ cm and shorter side $6$ cm, so the cone height is $6\sqrt3$ cm. Hence the volume is $\frac13\pi(6^2)(6\sqrt3)=72\sqrt3\pi\text{ cm}^3$ .
5	$\left(102-6\sqrt3\right)\text{ cm}^2$	4	Dropping perpendiculars from $D$ and $C$ to $AB$ creates special right triangles. At $A$ , the horizontal offset is $6/\tan60^\circ=2\sqrt3$ cm. At $B$ , the offset is $6/\tan45^\circ=6$ cm. Thus $CD=20-2\sqrt3-6=14-2\sqrt3$ cm. The area is $\frac12(20+14-2\sqrt3)(6)=102-6\sqrt3\text{ cm}^2$ .

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

Tier 1 · Easy

Mark scheme for G9 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$\tan x=\frac34$	2	$\cos^2x=1-\sin^2x=1-9/25=16/25$ . Since $x$ is acute, $\cos x=4/5$ . Hence $\tan x=(3/5)/(4/5)=3/4$ .
2	$2$	2	First, $\sin^2x+\cos^2x=1$ . Also $\tan x=\sin x/\cos x$ , so $\frac{\tan x\cos x}{\sin x}=1$ . The expression therefore simplifies to $1+1=2$ .

Tier 2 · Standard

Mark scheme for G9 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$\sin x$	2	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.
2	$\sin x\cos x=-\frac38$	3	Square the given equation: $(\sin x+\cos x)^2=1/4$ . Expanding and using $\sin^2x+\cos^2x=1$ gives $1+2\sin x\cos x=1/4$ . Hence $2\sin x\cos x=-3/4$ , so $\sin x\cos x=-3/8$ .
3	$\sin^2x=\frac45$ $\cos^2x=\frac15$	3	Since $\tan^2x=\sin^2x/\cos^2x$ , $\sin^2x=4\cos^2x$ . Substituting into $\sin^2x+\cos^2x=1$ gives $5\cos^2x=1$ . Hence $\cos^2x=1/5$ and $\sin^2x=4/5$ .

Tier 3 · Hard

Mark scheme for G9 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x}=\frac{2}{\sin x}$	4	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.
2	$(\sin x+\cos x)^4+(\sin x-\cos x)^4=2+8\sin^2x\cos^2x$	4	Let $s=\sin x$ and $c=\cos x$ . Expanding and adding gives $(s+c)^4+(s-c)^4=2s^4+12s^2c^2+2c^4=2(s^4+c^4+6s^2c^2)$ . Since $s^2+c^2=1$ , $s^4+c^4=(s^2+c^2)^2-2s^2c^2=1-2s^2c^2$ . Substitution gives $2(1+4s^2c^2)=2+8s^2c^2$ , as required.
3	$\frac{\tan x+\sin x}{\tan x-\sin x}=\frac{1+\cos x}{1-\cos x}=\frac{(1+\cos x)^2}{\sin^2x}=\left(\frac{1+\cos x}{\sin x}\right)^2$	4	Replace $\tan x$ by $\sin x/\cos x$ . The left side becomes $\frac{\sin x(1+\cos x)/\cos x}{\sin x(1-\cos x)/\cos x}=\frac{1+\cos x}{1-\cos x}$ . Multiplying numerator and denominator by $1+\cos x$ gives $\frac{(1+\cos x)^2}{1-\cos^2x}$ . Since $1-\cos^2x=\sin^2x$ , this is $\left(\frac{1+\cos x}{\sin x}\right)^2$ .
4	$\sin x=\frac45$ , $\cos x=-\frac35$ and $\tan x=-\frac43$	4	Squaring the given equation gives $\sin^2x+\cos^2x-2\sin x\cos x=49/25$ , so $\sin x\cos x=-12/25$ . Hence $(\sin x+\cos x)^2=1+2\sin x\cos x=1/25$ . In the stated interval $\sin x+\cos x>0$ , so it equals $1/5$ . Solving the simultaneous equations $\sin x-\cos x=7/5$ and $\sin x+\cos x=1/5$ gives $\sin x=4/5$ and $\cos x=-3/5$ , so $\tan x=-4/3$ .
5	$\frac{2\sqrt{10}}5$	3	Using $\tan x=\sin x/\cos x$ , the given expression becomes $\frac{\sin^2x+\cos^2x}{\sin x\cos x}=\frac1{\sin x\cos x}$ . Therefore $\sin x\cos x=3/10$ . It follows that $(\sin x+\cos x)^2=1+2\sin x\cos x=8/5$ . Both sine and cosine are positive for an acute angle, so $\sin x+\cos x=\sqrt{8/5}=2\sqrt{10}/5$ .

G10 · Solution of simple trigonometric equations in given intervals

Tier 1 · Easy

Mark scheme for G10 Tier 1 · Easy
Q	Answer	Mark	Comments
1	$x=90^\circ$	1	The sine graph reaches its maximum value $1$ at $90^\circ$ once in the interval, so $x=90^\circ$ .
2	$x=0^\circ$ or $x=180^\circ$	2	Tangent is zero at multiples of $180^\circ$ . The multiples in the interval are $0^\circ$ and $180^\circ$ ; $360^\circ$ is excluded.

Tier 2 · Standard

Mark scheme for G10 Tier 2 · Standard
Q	Answer	Mark	Comments
1	$x=150^\circ$ or $x=210^\circ$	2	The reference angle is $30^\circ$ . Cosine is negative in quadrants II and III, giving $x=180^\circ-30^\circ=150^\circ$ and $x=180^\circ+30^\circ=210^\circ$ .
2	$x=60^\circ$ or $x=240^\circ$	3	The reference angle is $60^\circ$ because $\tan60^\circ=\sqrt3$ . Tangent has period $180^\circ$ , so the solutions are $x=60^\circ+180^\circ n$ . The values in the interval are $60^\circ$ and $240^\circ$ .
3	$x=225^\circ$ or $x=315^\circ$	3	The reference angle is $45^\circ$ because $\sin45^\circ=\frac{\sqrt2}{2}$ . Sine is negative in the third and fourth quadrants, so the solutions are $180^\circ+45^\circ=225^\circ$ and $360^\circ-45^\circ=315^\circ$ .

Tier 3 · Hard

Mark scheme for G10 Tier 3 · Hard
Q	Answer	Mark	Comments
1	$x=60^\circ,300^\circ$	4	Use $\sin^2x=1-\cos^2x$ . Writing $c=\cos x$ gives $2(1-c^2)=3c$ , so $2c^2+3c-2=0=(2c-1)(c+2)$ . Hence $c=1/2$ or $c=-2$ . The second value is impossible for a cosine, while $\cos x=1/2$ gives $x=60^\circ$ or $300^\circ$ in the stated interval.
2	$k=1$ gives $x=90^\circ$ ; $k=-1$ gives $x=270^\circ$	4	If $\|k\|>1$ there is no solution. If $-1<k<0$ or $0<k<1$ , the horizontal line at height $k$ meets one sine cycle twice. For $k=0$ , the included endpoints give three solutions: $0^\circ$ , $180^\circ$ and $360^\circ$ . Only the maximum and minimum levels meet the sine curve once, so $k=1$ gives $x=90^\circ$ and $k=-1$ gives $x=270^\circ$ .
3	$x=60^\circ,120^\circ,240^\circ,300^\circ$	4	A solution cannot have $\cos x=0$ , because the equation would then force $\sin^2x=0$ as well, contradicting $\sin^2x+\cos^2x=1$ . So divide by $\cos^2x$ to obtain $\tan^2x=3$ . Hence $\tan x=\sqrt3$ or $\tan x=-\sqrt3$ , with reference angle $60^\circ$ . Using all four quadrants gives $x=60^\circ,120^\circ,240^\circ,300^\circ$ .
4	$x=126.9^\circ$ or $x=306.9^\circ$	3	The reference angle is $\tan^{-1}(4/3)=53.130\ldots^\circ$ . Tangent is negative in quadrants II and IV, so the solutions are $180^\circ-53.130\ldots^\circ=126.869\ldots^\circ$ and $360^\circ-53.130\ldots^\circ=306.869\ldots^\circ$ . These round to $126.9^\circ$ and $306.9^\circ$ .
5	$x=90^\circ$	4	The original denominator requires $1+\cos x\ne0$ . Rearranging gives $\sin x=1+\cos x$ . Squaring and using $\sin^2x=1-\cos^2x$ gives $1-\cos^2x=1+2\cos x+\cos^2x$ , so $2\cos x(\cos x+1)=0$ . The candidates are $\cos x=0$ and $\cos x=-1$ . Checking in the original equation keeps $x=90^\circ$ , rejects $270^\circ$ , and excludes $180^\circ$ because the denominator is zero. Therefore $x=90^\circ$ .