HookHow they measured Everest without touching it
On 9 November 1852, in the survey office at Dehradun in northern India, the chief computer of the Great Trigonometrical Survey, Radhanath Sikdar, is said to have hurried in to the Surveyor-General, Sir Andrew Waugh, with the words: 'Sir, I have discovered the highest mountain in the world.' Nobody had climbed it. Nobody had stood within a hundred miles of its base with a tape. Its height — announced in 1856 as 29,002 feet, and later named Everest — had been worked out from angles alone, measured with theodolites from six survey stations out on the plains, some of them more than 100 miles from the peak. The famous extra two feet were reportedly added on top of a suspiciously round 29,000 so the figure would not look like a guess.
That is the quiet power of this whole section: measurement lets you pin down a length or an angle you could never reach, using quantities you can. This is where the formulae live and where the calculator paper earns most of its marks. You will meet standard units and the conversions that trip people up; the areas of triangles, parallelograms and trapezia and the volumes of prisms and cylinders; circles, arcs and sectors; the surface area and volume of spheres, cones and composite solids; Pythagoras' theorem and trigonometry in both two and three dimensions; the exact trigonometric values you must know without a calculator; and finally the two rules that unlocked Everest — the sine rule and the cosine rule for any triangle, together with the area formula \(\tfrac{1}{2}ab\sin C\) and the way lengths, areas and volumes scale in similar figures.
ModelUnits, measuring, scale and bearings
The metric backbone is assumed throughout Higher: millimetres, centimetres, metres and kilometres for length; grams, kilograms and tonnes for mass; millilitres and litres for capacity; seconds, minutes and hours for time; and the compound units built from them — speed in metres per second, density in grams per cubic centimetre, pressure in newtons per square metre. The one conversion that separates grades is the one that squares or cubes. Because \(1\text{ m}=100\text{ cm}\), a square metre is \(100\times 100=10{,}000\) cm\(^2\) and a cubic metre is \(100\times 100\times 100=1{,}000{,}000\) cm\(^3\). Convert an area and you square the factor; convert a volume and you cube it. Miss that and your answer is out by a hundred or a thousand.
Measuring is the practical half. Line segments are read with a ruler and angles with a protractor, but Higher leans hardest on bearings and scale drawings. A bearing is always measured from north, turned clockwise, and written with three figures, so due east is \(090^\circ\) and a back-bearing differs from the forward bearing by exactly \(180^\circ\). A map scale such as \(1:25{,}000\) means one unit on the paper is 25,000 of the same units on the ground, so 1 cm represents 250 m. These conventions feed straight into the trigonometry and vector work later, where a bearing fixes a direction and a triangle supplies the distance.
A footpath measures \(8\) cm on a \(1:25{,}000\) map, running out from a car park on a bearing of \(065^\circ\). How long is it on the ground, and what is the bearing of the walk back? The real length is \(8\times 25{,}000=200{,}000\) cm \(=2{,}000\) m \(=2\) km. The return bearing is the forward bearing plus \(180^\circ\): \(065^\circ+180^\circ=245^\circ\). Multiply by the scale for the distance, add \(180^\circ\) for the back-bearing — and keep the leading zero on \(065^\circ\), because the three-figure form is part of the mark.
MechanismArea and volume of the everyday shapes
Three area formulae must be recalled from memory. A parallelogram is \(\text{base}\times\text{perpendicular height}\); a triangle is \(\tfrac{1}{2}\times\text{base}\times\text{perpendicular height}\), because any triangle is exactly half a parallelogram on the same base; and a trapezium is \(\tfrac{1}{2}(a+b)h\), the average of its two parallel sides multiplied by the perpendicular gap between them. In every case the height is the straight-across perpendicular height, never the slanted edge — using the slant side is the most common area slip in the topic.
Volume steps up to three dimensions through the idea of a prism: any solid with the same cross-section all the way along. Its volume is \(\text{area of cross-section}\times\text{length}\), which is why a cuboid is \(l\times w\times h\) (a rectangular cross-section) and a cylinder is \(\pi r^2\times h\) (a circular one). Find the cross-sectional area first, then multiply by the length; naming the cross-section is what turns an intimidating solid into a clean two-step calculation.
A water trough has a trapezium cross-section with parallel sides \(40\) cm and \(60\) cm and a perpendicular depth of \(30\) cm, and it is \(200\) cm long. How many litres does it hold? The cross-section area is \(\tfrac{1}{2}(40+60)\times 30=\tfrac{1}{2}\times 100\times 30=1{,}500\) cm\(^2\). The trough is a prism, so its volume is \(1{,}500\times 200=300{,}000\) cm\(^3\). Since \(1{,}000\) cm\(^3=1\) litre, that is \(300\) litres. Cross-section area, then length — the same plan works for any prism.
ModelCircles, sectors and the curved solids
Two circle formulae are not on the AQA formula sheet and must be memorised: the circumference \(C=\pi d=2\pi r\) (a length, radius used once) and the area \(A=\pi r^2\) (a region, radius squared). A sector is a slice of a circle and simply takes the same fraction of each. For a centre angle \(\theta\), the arc length is \(\tfrac{\theta}{360}\times 2\pi r\) and the sector area is \(\tfrac{\theta}{360}\times\pi r^2\). A \(150^\circ\) sector of radius \(12\) cm therefore has arc \(\tfrac{150}{360}\times 24\pi=10\pi\approx 31.4\) cm and area \(\tfrac{150}{360}\times 144\pi=60\pi\approx 188.5\) cm\(^2\): find the fraction once, then take that share of the circumference or the area.
Higher adds the curved solids. The formula sheet supplies the sphere (volume \(\tfrac{4}{3}\pi r^3\), surface area \(4\pi r^2\)) and the cone (volume \(\tfrac{1}{3}\pi r^2 h\), curved surface \(\pi r l\)), leaving you to substitute carefully and, for a composite solid, to add the parts while ignoring any face buried inside the join. Keep volumes in cubed units and areas in squared units, and leave answers in terms of \(\pi\) whenever an exact value is asked for.
A toy is a cone of radius \(3\) cm and slant height \(5\) cm sitting on a hemisphere of radius \(3\) cm. Find its exact volume and outer surface area. The cone's vertical height is \(\sqrt{5^2-3^2}=\sqrt{16}=4\) cm, so its volume is \(\tfrac{1}{3}\pi\times 3^2\times 4=12\pi\). The hemisphere is half a sphere: \(\tfrac{1}{2}\times\tfrac{4}{3}\pi\times 3^3=18\pi\). Total volume \(=12\pi+18\pi=30\pi\approx 94.2\) cm\(^3\). For the surface, count only the outer skin: the cone's curved surface \(\pi r l=\pi\times 3\times 5=15\pi\) plus the hemisphere's curved surface \(2\pi r^2=2\pi\times 9=18\pi\), giving \(33\pi\approx 103.7\) cm\(^2\). The flat circle where the two meet is inside the solid, so it is not part of the surface.
MechanismPythagoras and trigonometry, in 2D and 3D
In a right-angled triangle, Pythagoras' theorem \(a^2+b^2=c^2\) links the three sides, with \(c\) the hypotenuse opposite the right angle. Add the squares of the two shorter sides to find the hypotenuse; subtract to find a shorter side. Trigonometry then ties a ratio of sides to an angle through SOH-CAH-TOA: \(\sin\theta=\tfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\tfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta=\tfrac{\text{opp}}{\text{adj}}\). Choose the ratio that uses your two sides, then rearrange for a missing length or use the inverse buttons \(\sin^{-1},\ \cos^{-1},\ \tan^{-1}\) for a missing angle.
Higher pushes both into three dimensions, and the method never changes: find a right-angled triangle inside the solid, usually by working out a diagonal across a base with Pythagoras first, then using that diagonal as one side of a vertical triangle. Higher also expects the exact values without a calculator: \(\sin 30^\circ=\cos 60^\circ=\tfrac{1}{2}\), \(\sin 60^\circ=\cos 30^\circ=\tfrac{\sqrt{3}}{2}\), \(\sin 45^\circ=\cos 45^\circ=\tfrac{\sqrt{2}}{2}\), \(\tan 45^\circ=1\), with \(\sin 0^\circ=0\) and \(\cos 0^\circ=1\). These surface on Paper 1, the non-calculator paper, precisely because you cannot look them up there.
A room is a cuboid \(6\) m long, \(4\) m wide and \(3\) m high. Find the length of the longest straight rod that fits corner to opposite corner, and the angle it makes with the floor. First the floor diagonal, by Pythagoras: \(\sqrt{6^2+4^2}=\sqrt{52}\approx 7.21\) m. That diagonal and the vertical height form a right-angled triangle, so the space diagonal is \(\sqrt{52+3^2}=\sqrt{61}\approx 7.81\) m. The angle \(\theta\) to the floor satisfies \(\tan\theta=\tfrac{3}{\sqrt{52}}=\tfrac{3}{7.21}=0.4160\), so \(\theta=\tan^{-1}(0.4160)\approx 22.6^\circ\). Base diagonal first, vertical triangle second — that two-stage plan solves every 3D trigonometry problem.
CaseAny triangle at all — the sine and cosine rules
Everest lay in no right-angled triangle, and neither do most real ones, so SOH-CAH-TOA runs out. Two rules handle any triangle. The sine rule \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\) is for when you have a matching side-and-angle pair plus one more fact; it is exactly how the Survey of India turned a measured baseline and two theodolite angles into the distance to a peak. The cosine rule \(c^2=a^2+b^2-2ab\cos C\) starts the two cases the sine rule cannot: three sides given (find an angle) or two sides and the angle between them (find the third side).
The third tool finds area with no perpendicular height at all: \(\text{Area}=\tfrac{1}{2}ab\sin C\), where \(C\) is the angle between the two sides \(a\) and \(b\). All three are printed on the AQA formula sheet, so the marks are for choosing correctly — cosine rule for SSS or SAS, sine rule otherwise — and for watching the sine rule's ambiguous case, where finding an angle can give an obtuse partner \(180^\circ-\theta\) that also fits the triangle.
A triangular plot has two sides of \(8\) m and \(5\) m with an included angle of \(60^\circ\) between them. Find the third side and the area. The third side faces the known angle, so use the cosine rule: \(c^2=8^2+5^2-2\times 8\times 5\times\cos 60^\circ=64+25-80\times\tfrac{1}{2}=89-40=49\), giving \(c=\sqrt{49}=7\) m — a whole number, the sign of a well-set problem. For the area, use \(\tfrac{1}{2}ab\sin C=\tfrac{1}{2}\times 8\times 5\times\sin 60^\circ=20\times\tfrac{\sqrt{3}}{2}=10\sqrt{3}\approx 17.3\) m\(^2\). The exact value \(\sin 60^\circ=\tfrac{\sqrt{3}}{2}\) let the area stay exact as \(10\sqrt{3}\); reaching for the calculator too early would throw that exactness away.
DataSimilar figures — why area and volume scale differently
Two figures are similar when one is a scaled copy of the other: equal angles, and every length multiplied by the same length scale factor \(k\). The Higher headline is what then happens to areas and volumes. Because area is a length squared, the area scale factor is \(k^2\); because volume is a length cubed, the volume scale factor is \(k^3\). Double every length and the surface area quadruples while the volume grows eightfold — the same squaring and cubing that ran through the units at the start of this section, and the reason an 18-inch pizza dwarfs a 12-inch one.
This works in reverse, and that is where marks hide. Given a ratio of areas, the length ratio is its square root; given a ratio of volumes — or of masses, since for one material mass is proportional to volume — the length ratio is its cube root. Congruent figures are just the special case \(k=1\), identical in size as well as shape, decided by the criteria SSS, SAS, ASA and RHS, whereas equal angles alone are enough for similarity.
Two solid bronze statues are mathematically similar. The smaller is \(20\) cm tall, needs \(500\) cm\(^2\) of gold leaf and has a mass of \(2\) kg; the larger is \(60\) cm tall. Find the leaf and the mass for the larger. The length scale factor is \(k=\tfrac{60}{20}=3\). Area scales by \(k^2=9\), so the leaf needed is \(500\times 9=4{,}500\) cm\(^2\). Mass follows volume, which scales by \(k^3=27\), so the mass is \(2\times 27=54\) kg. Length triples, area is \(\times 9\), volume and mass are \(\times 27\) — miss the squaring or cubing and the answer is out by a large factor.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
This section is the heartland of the calculator papers (Papers 2 and 3), where a circle, a volume, a Pythagoras or a trigonometry problem is usually worth several marks each. Know the formula sheet cold before you walk in: AQA gives you the sine rule, the cosine rule, the area \(\tfrac{1}{2}ab\sin C\), the prism volume, and the sphere and cone formulae — but not the circumference or area of a circle, which you must still recall from memory.
Mark schemes here are method-first. Write the formula, substitute the numbers, then evaluate — a correct substitution with a slipped final digit usually keeps most of the marks, while a bare wrong answer keeps none. Keep full calculator accuracy all the way through and round only on the final line; using 3.14 for \(\pi\) or 0.87 for \(\sin 60^\circ\) mid-calculation can lose the accuracy mark. When a question says 'in terms of \(\pi\)' or 'give an exact answer', leave \(\pi\) and surds in — a decimal there loses the mark even on a calculator paper.
For any non-right-angled triangle, decide the rule first: three sides or two-sides-and-the-angle-between means cosine rule, otherwise sine rule, and always check the sine rule's ambiguous case when the missing angle could be obtuse. For 3D problems, draw the internal right-angled triangle and solve the base diagonal before the vertical one. Match units to the dimension — cm for length, cm\(^2\) for area, cm\(^3\) for volume — and on Paper 1 expect exact values and numbers that come out whole.