Question Working Answer Mark AO Notes Grade 1 a i ii iii iv b i ii iii iv c No Yes Yes No Assume third angle is 90°. 180° – 90° = 90°. So both the remaining angles must be acute. If the third angle is bigger than 90° both remaining must also be acute. If the third angle is acute you would need to make one of the other angles at least 90°. Example with 2 acute angles. Example with 1 obtuse angle. The obtuse angles will be (90° + x) and (90° + y), adding these together you get 180° + x + y, which is more than the sum of the angles in a triangle, so it’s impossible. If you draw a line between two parallel lines, the two allied angles formed add up to 180°, which leaves nothing for a third angle. B1 B1 B1 B1 C1 C1 B1 B1 C1 C1 C1 C1 2 C1 for full explanation C1 for clarity of explanation B1 for giving an example that works B1 for giving an example that works C1 for a correct explanation C1 for clear communication C1 for clear explanation C1 for clarity of the communication B 12
10 a b Angles in a triangle add up to 180°. You can split any quadrilateral into two triangles. Therefore, the sum of the interior angles of any quadrilateral = 2 × 180o. For the two outside triangles, use the sum of the angles in triangle = 180° and the interior angle of a regular pentagon = 108°. 108° + 2x = 180° 2x = 180° – 108° 2x = 72° x = 36° For the middle triangle, use interior angle of regular pentagon = 108°. y = 108° – x y = 108° – 36° y = 72° Using the sum of angles in triangle = 180°: z = 180 – 2y z = 180 – 144 z = 36o Or 2x + z = 108° x = 36° so 2 × 36° + z = 108° z + 72° = 108° z = 36° Two triangles with one angle = 108° and two other equal angles of 36°. One triangle with one angle = 36° and two angles = 72°. C1 C1 B1 C1 C1 C1 M1 A1 M1 A1 M1 A1 C1 C1 2 C1 for clear explanation C1 for communication with clear diagram B1 for showing inter angles of quadrilateral = 2 × 180 C1 for a clear diagram correctly showing the three triangles C1 for identifying each angle of the triangles with symbols showing which angles are equal C1 for clear explanation M1 for correct method of finding angle x A1 for 36° M1 for correct method of finding angle y A1 for 72° M1 for correct method of finding angle z A1 for 36° C1 for clear argument and stating assumptions used C1 for use of diagram with clarity of explanation M 14
25 Using trigonometry: cos α = α = cos–1 α = 33.557 31° Area of segment of one circle = πr2 – r2sinθ = r2(π – sin θ) where θ = 2 × 33.6° = 67.2° and r = 6 cm. Area = 62(π –sin67.2°) = 36(0.586 431 –sin 67.2°) = 36(0.586 431 – 0.460 931 6) = 4.517 979 cm2 Area of overlap = 2 × 4.517 979 cm2 = 9.035 959 cm2 9.0 cm2 M1 A1 M1 M1 A1 A1 M1 A1 C1 2 M1 for correct trigonometric statement for angle A1 for any rounded answer to 2 or more sf M1 for correct segment formula M1 for correct factorisation A1 for correct substitution of radius and a correct angle A1 for correct answer to 2 or more sf M1 for multiplying by 2 A1 for correct answer to either 1 or 2 sf C1 for use of mathematical language and diagrams to support solution M 9
29 Let c = the height of the chimney. = tan 53° x = c tan 53° = tan 62° 30 + x = c tan 62° x = c tan 62° – 30 Combining equations 1 and 2 to eliminate x: c tan 53° = c tan 62° – 30 Rearrange to get c on one side of the equation. 30 = c tan 62° – c tan 53° 30 = c(tan 62° – tan 53°) c = = 54.182 761 m 54.2 m C1 M1 A1 M1 A1 M1 M1 A1 2 C1 for clear correct diagram used M1 for correct use of trigonometry with x, c and angle 53° or 37° A1 for correct equation having x as subject M1 for correct use of trigonometry with x, c and angle 62° or 28° A1 for correct equation in format to combine with equation 1 M1 for correctly eliminating x M1 for correct equation with c as subject A1 for correct rounding to 2 or 3 sf M 8
30 Angle at A is 90° + (360° – 330°) = 120° Angle at B is 290° – 270° = 20° Angle at S is 180° – (120° + 20°) = 40° Use the sine rule. = x = 15 × = 7.981 33 8.0 km C2 B1 B1 B1 M1 A1 2 C1 for diagram illustrating how angles found at A and B C1 for complete triangle drawn, showing all relevant data B1 for 120° B1 for 20° B1 for 40° M1 for use of sine rule A1 for correct answer rounded to 1, 2 or 3 sf M 7 31 a b c Sometimes true. It is not true if the number of individual cubes has fewer than 3 factors, including 1 and itself, for example, you cannot do it with 7 cubes (factors 1 and 7) You can only make one cuboid if the number of factors, including 1 is equal to 3, for example, with 21 cubes (factors 1, 3 and 7). You can make more than one cuboid if the number of cubes has more than 3 factors not including itself, for example, 30 (factors 1, 2, 3 and 5). C1 C1 C1 C1 C1 2 C1 for sometimes true C1 for clear explanation for when not true C1 for clear explanation for when only 1 cuboid could be made C1 for clear explanation for when more than 1 cuboid could be made C1 for use of examples to illustrate the explanations M 5
34 Consider just half the shape, where x is the length of the string. Use Pythagoras’ theorem. x2 = 102 + 22.52 = 606.25 x = = 24.622 145 Two lengths of string will be 49.244 289 cm Subtract the original 45 cm Gives extension as 4.244 289 4.2 cm C1 M1 M1 A1 A1 A1 2 C1 for clear diagram M1 for correct Pythagoras statement M1 for correct method of applying Pythagoras’ theorem A1 for full answer A1 for double the initial x A1 for rounded answer to either 2 or 3 sf M 6 35 Let AC = x, the length of the new road. Use Pythagoras’ theorem. x2 = 4.92 + 6.32 = 63.7 x = = 7.981 228 Current distance = 4.9 + 6.3 = 11.2 km Saving = 11.2 – 7.981 228 = 3.218 772 km 3.22 km C1 M1 M1 A1 B1 M1 A1 2 C1 for use of a diagram to assist the explanation M1 for clear statement of Pythagoras M1 for correctly applying Pythagoras’ theorem A1 for full answer B1 for 11.2 M1 for subtracting lengths A1 for correct rounding to 2 or 3 sf M 7 36 Yes. = sin–1 = 53.13° = 53° to the nearest degree = 50° to 1 sf 12 cm has range of 11.5 to 12.5 cm 15 cm has range of 14.5 to 15.5 cm The smallest value for sin is which gives = sin–1 0.7419 = 47.9° So there are values that round to 12 cm and 15 cm which will give an angle that rounds to 50°. B1 P1 P1 B1 C1 2 B1 for yes P1 for showing that using trigonometry and rounding can give 50° P1 for showing the ranges of lengths of the sides B1 for showing the least possible value of the angle given the ranges C1 for final summary explaining that it is possible M 5
37 Use Pythagoras’ theorem. AC2 = 42 – (2)2 = 16 – 8 = 8 BC2 = 8 = (2)2 = 8 Hence BC = AC, an isosceles triangle. M1 A1 M1 A1 C1 2 M1 for correct Pythagoras statement A1 for correct value of AC2 M1 for finding BC2 A1 for correct value of BC2 C1 for clear explanation of sides being the same length M 5 38 AB2 = 22 – 12 = 4 – 1 = 3 AB = cm M1 A1 C1 2 M1 for correct Pythagoras statement A1 for 3 C1 for a clear communication of the method used M 3 39 cos 68° = –cos 112° = –cos 248° = 0.3746 cos 338° = 0.9271 cos 338° is the odd one out. All the others have the same numerical value (ignoring signs). B1 C1 2 B1 for cos 338° C1 for a clear explanation M 2 40 a i ii b sin x + 1 = 2 sin x = 1 x = sin–11 = 90° 2 + 3cos x = 1 3cos x = 1 – 2 = –1 cos x = – x = 109.5° and 360° – 109.5° = 250.5° cos 320° = 0.766 044 4 sin–1 0.766 044 4 = 50° and 180° – 50° = 130° x = 90° x = 109.5°, 250.5° x = 50° and 130° M1 A1 M1 A1 A1 A1 M1 A1 A1 2 M1 for sin x = 1 A1 for 90° M1 for first step of solving equation A1 for correct statement of cos x A1 for correct angle to 1 dp A1 for correct angle to 1dp M1 for method of getting to sin–1 A1 for 50° A1 for 130° M 9
41 a b tan x = sin x ÷ cos x ÷ = = = = 1 tan–1 1 = 45° Alternatively, use Pythagoras’ theorem to work out the length of the third side, e.g. (side)2 = 16 – 8 = 8 side = tan x = = 1 tan-1 1 = 45° tan x = 1 45° C1 M1 A1 A1 M1 A1 2 C1 for communicating effectively how the sine and cos can be used to find tan (This could be tan x = ) M1 for correct use of tan A1 for correct combination of surds A1 for tan x = 1 M1 for use of inverse tan or recognising an isosceles triangle A1 for 45° M 6 42 Using Pythagoras’ theorem, the hypotenuse = = = 4 Then sin x = and cos x = Hence (sin x)2 + (cos x)2 = + = = 1 M1 A1 B1 B1 C1 2 M1 for correct Pythagoras statement A1 for 4 B1 for correct sin x B1 for correct cos x C1 for correct explanation M 5
43 Use the cosine rule. c2 = a2 + b2 - 2ab cos C = 9 + 16 – 2 × 3 × 4 × cos 60° = 25 – 24 × = 13 c = cm C1 M1 A1 B1 M1 A1 2 C1 for clear use of diagram M1 for correct cosine rule statement A1 for correct substitution B1 for cos 60° = M1 for taking square root A1 for correct surd form M 6 44 B1 B1 2 B1 for correct diagram B1 for correct vector M 2 45 No. To work out the return vector, multiply each component by –1. The return vector is . B1 B1 C1 2 B1 for No B1 for correct vector C1 for a clear explanation of what Joel should have done M 3
46 a b C3 C3 2 C1 for showing construction of 60° C1 for showing bisection of 60° C1 for showing bisection of 30° to show 15° C1 for showing construction of 90° C1 for showing construction of 60° to leave a 30° angle C1 for bisecting that 30° angle to leave 75° H 6 47 Each angle bisector is the locus of points equidistant from the two arms or sides of the angle. Hence, where they all meet will be the only point that is equidistant from each of the three sides. Hence a circle can be drawn inside the triangle, with this centre, that just touches each side of the triangle. C2 P1 2 M1 for explanation referring to the bisector being equidistant M1 for the interpretation of the intersections P1 for the use of a diagram to help interpret the explanation H 3
48 a–d e f A circle, centre P, should pass through each of the nine labelled points. M1 A1 M1 A1 A1 M1 A1 M1 A1 C1 2 M1 for method of constructing any midpoint A1 for all three midpoints correct and labelled M1 for constructing any perpendicular from a vertex to opposite face A1 for all three correct A1 all 3 feet correctly labelled M1 for constructing a midpoint of AO or BO or CO A1 for all midpoints correct and labelled M1 for constructing a bisector of LM or LN or MN A1 for all three bisectors correct and point of intersection labelled P C1 for correct explanation H 10 49 AB = CD (given) ∠ABD = ∠CDB (alternate angles) ∠BAC = ∠DCA (alternate angles) so ΔABX ≡ ΔCDX (ASA) C3 C1 2 C1 for correct statement with justification C1 for correct statement with justification C1 for correct statement with justification C1 for stating ASA within correct explanation H 4 50 AB and PQ are the corresponding sides opposite the 50° angle but they are not equal in length. C2 2 C1 for stating the corresponding side link C1 for complete clear statement H 2
51 a b Area of original rectangle = x2 If the length of the side of the inscribed square is h, then: h2 = h2 = h2 = Area of inscribed square is h2 = Joining adjacent midpoints forms a right-angled isosceles triangle, with the perpendicular sides each half the length of the side of the square. All four triangles are congruent. At each midpoint, there are two angles of 45°, leaving the vertex of the new shape being 90°. Since all the sides are equal and all the angles are 90°< the new shape is a square. The area of the inscribed square is half the area of the original square. C2 C1 M1 A1 A1 C1 2 C1 for explaining all sides are the same length C1 for explaining how all angles are 90° C1 for communicating a given original side length and finding the length of a side of the inscribed square M1 for correctly using Pythagoras’ theorem to find the length of the side of the new square A1 for correct expression of h2 A1 for correctly stating area C1 for clear communication of complete solution H 7 52 Each internal angle of an octagon is 135°. Each internal angle of a hexagon is 120°. The sum of these two angles is 255°. The sum of the angles in a quadrilateral is 360° so the sum of the remaining angles is: 360° – 255° = 105°. The two remaining angles are equal, as the line joining through the vertices C, J, G and L (and thus through the obtuse angles) is a line of symmetry. JFG = 52.5° C2 B1 M1 A1 C2 B1 2 C1 for explanation of 135° C1 for explanation of 120° B1 for 255° M1 for subtraction from 360° A1 for 105° C1 for explaining two angles are equal C1 for clear reasons given as to why B1 for correct 52.5° H 8
53 Use the sine rule in both triangles. As A = P, then sin P = sin A So Hence = Hence = So p = M1 A1 A1 M1 B1 P1 2 M1 for use of sine rule A1 for correct equation A1 for correct equation M1 for equating both known angles B1 for correct statement linking p, a, b and c P1 for clear communication of the full solution H 6
54 Use Pythagoras’ theorem to write expressions for z2. Triangle ABC: z2 = r2 + (s + y)2 Triangle ABD: z2 = w2 + (x + t)2 So r2 + (s + y)2 = w2 + (x + t)2 Now r2 = x2 – s2 and w2 = y2 – t2 Substitute for r2 and w2 in the equation: x2 – s2 + (s + y)2 = y2 – t2 + (x + t)2 Multiply out the brackets: x2 – s2 + s2 + 2sy + y2 = y2 – t2 + x2 + 2xt + t2 Simplify: x2 + 2sy + y2 = y2 + x2 + 2xt 2sy = 2xt sy = xt Alternatively: Assume T is the intersection of AD and BC. ACT is similar to BDT as both contain the same angles. (Angles ATC and BTD are vertically opposite and so are equal.) Hence by similarity = So xt = sy P1 P1 P1 P1 P1 C1 C1 2 P1 for combining expressions P1 for substitution P1 for multiplying out to get individual terms P1 for simplifying P1 for clear explanation C1 for = C1 for clear communication of proof H 7
= 2x2 + t(m + n + t) But m + n + t = x So A = 2x2 + tx But the sides are of length 2x, so A = 4x2 Then 2x2 + tx= 4x2 and tx= 2x2 t= 2x Squaring each side gives: 5t2 = 4x2 t2 = Whole area A = 4x2 Hence the middle square, t2, is of A. M1 A1 M1 B1 M1 A1 M1 A1 C1 C1 A1 for (m + n)t M1 for adding all separate components A1 for correct statement of A M1 for substituting (m + n + t) B1 for whole area 4x2 M1 for equating the two equations A1 for equation enabling t to be identified M1 for squaring A1 for correct expression for t2 C1 for clear explanation showing idea C1 for complete clear solution 15
64 Draw triangle BCP, where P is on the circumference in the opposite segment . Then angle BPC is 180° – x as BAC and BPC are opposite angles in a cyclic quadrilateral. Angle BOC is double BPC, which is 360° – 2x. Angle BCO will also be y, as BCO is an isosceles triangle. Angles in triangle OBC sum to 180°, so 2y + 360° – 2x = 180° and 2y = 2x – 180° Divide both sides by 2 to give: y = x – 90°. C1 M1 A1 A1 B1 M1 A1 C1 2 C1 for clear use of diagram M1 for using point P A1 for 180 – x A1 for identifying 360° – 2x B1 for identifying BCO M1 for adding known angles together A1 for simplifying C1 for a clear, well presented solution H 8 65 As ABCD is a cyclic quadrilateral, the opposite angles will sum to 180°. So 2x – 5° + 5y – 20° = 180° 2x + 5y = 205° ………(1) And 3y + 5 + 2x + 20° = 180° 2x + 3y = 155° ……..(2) Subtract (2) from (1): 2y = 50° y = 25° Substitute y into (2): 2x + 75° = 155° 2x = 80° x = 40° y = 25° x = 40° C1 M1 A1 A1 M1 A1 M1 A1 2 C1 for explanation of cyclic quadrilateral M1 for adding opposite angles A1 for first correct equation A1 for second correct equation M1 for method of eliminating one variable A1 for 25° M1 for method of substitution A1 for 40° H 8
66 Drop perpendiculars down to Y and X and complete the trapezium, with the top triangle being ABT, as shown. Length TB will be the same as YZ. Use Pythagoras’ theorem on triangle ABT. TB2 = 92 – 32 = 81 – 9 = 72 TB = = 8.485 281 4 8.49 cm C1 C1 C1 M1 A1 A1 3 C1 for good clear diagram showing perpendiculars to Y and Z C1 for clear diagram of ABT C1 for clear explanation of how the triangle ABT has been formed M1 for use of Pythagoras’ theorem A1 for correct statement A1 for correct answer to 2 or 3 sf H 6 67 Circle Theorem 1 states The angle at the centre of a circle is twice the angle at the circumference when they are both subtended by the same arc.Theorem 1 tells you that any angle subtended from arc AB at a point C on the circumference is always half the angle subtended from arc AB at the centre. Therefore, every possible angle subtended from arc AB at the circumference will correspond to the same angle at the centre. Hence angles subtended at the circumference are equal. P2 C1 C1 3 P1 for stating Theorem 1 P1 for extending this to this proof C1 for clear overall explanation and clarity C1 for use of diagram and mathematical language. H 4
78 Using angles on a straight line. = 180° – 58° = 122° Angle ADB = = 180° – (32° + 122°) = 180° – 154° = 26° Use the sine rule. e = sin 32° × e = sin 32° × 63.872 816 9 = 63.847 436 Using trigonometry: sin 58° = c = e sin 58° = 28.704 253 7 So the height of the tower is 29 m to the nearest metre. M1 B1 M1 B1 M1 M1 A1 C1 3 M1 for method for finding angle ABD B1 for 26° M1 for use of sine rule B1 for e and any rounding M1 for correct use of trigonometry M1 for correct expression for c A1 for 29 answer rounded to 2 or 3 sf A1 for complete accurate solution well laid out. H 8
A1 C1 M1 for method of finding β A1 for β to 4 or more sf A1 for final three-figure bearing C1 for effective use of diagrams and good use of mathematical language 20 80 Use the sine rule to work out angle B. sin B = B = sin–1 = 62° But we know ABC is obtuse, so: B = 180° – 62° = 118° Then C = 180° – (118° + 32°) = 30° Use the sine area rule. Area = = × 10 × 6 × sin 30° = 15 cm2 15 cm2 M1 A1 A1 B1 M1 M1 A1 C1 3 M1 for use of sine rule A1 for 62°, accept any rounding to 2sf or more A1 for 118° and any rounding to 3sf or more B1 for 30° with any rounding to 2sf or more M1 for use of area sine rule M1 for correct substitution A1 for 15 rounded to 2 or 3 sf C1 for a complete solution clearly set out with correct mathematical language and symbols H 8 81 a Use Pythagoras’ theorem to work out H. H = = 26 Use the cosine rule to work out angle D. cos D = = 0.275 D = 74.037 986° Use the area sine rule to work out the area of triangle ADC. Area = × 20 × 23 × sin 74.037 986° = 221.13217 m2 Area of triangle ABC = × 24 × 10 = 120 m2 Total area = 341.132 17 m2 C1 M1 A1 M1 3 C1 for use of clear diagram M1 for use of Pythagoras’ theorem A1 for 26 cao M1 for use of cosine rule H