Mathematics 0580 | Paper 2 Mock 01 | Non-calculator (Extended)

(a) Two sides and the angle between them are given, which is exactly the situation for the cosine rule: \(a^{2}=b^{2}+c^{2}-2bc\cos A\), where \(A\) is the included angle and \(a\) the side opposite it.

1. \(BC^{2}=8^{2}+5^{2}-2\times 8\times 5\times\cos 60^{\circ}\) [M1].
2. Using the exact value \(\cos 60^{\circ}=\dfrac{1}{2}\), the last term is \(80\times\dfrac{1}{2}=40\), so \(BC^{2}=64+25-40=89-40=49\) [M1].
3. \(BC=\sqrt{49}=7\) cm [A1].

(b) With the same two sides and included angle, the area is \(\dfrac{1}{2}ab\sin C\):

Area \(=\dfrac{1}{2}\times 8\times 5\times\sin 60^{\circ}\) [M1] \(=20\times\dfrac{\sqrt{3}}{2}=10\sqrt{3}\) cm\(^{2}\) [A1] cao.

The same pair of sides and the same angle serve both parts, but through different ratios: cosine for the length, sine for the area. Mixing them up is the usual error.

Because \(\cos 60^{\circ}\) is exactly \(\dfrac{1}{2}\), the cosine rule reduces here to \(BC^{2}=b^{2}+c^{2}-bc\), which is why the answer comes out as a whole number without a calculator.

This is a reverse percentage. The \(20\%\) was taken off the original price, so \(\$68\) represents \(100\% - 20\% = 80\%\) of the original, not \(100\%\).

If the original price is \(x\), then \(0.8x = 68\), so \(x = 68 \div 0.8\) [M1] oe.

Clear the decimal: \(\dfrac{68}{0.8} = \dfrac{680}{8} = 85\), so the original price was \(\$85\) [A1].

Check: \(20\%\) of \(\$85\) is \(\$17\), and \(85 - 17 = 68\), which matches the sale price.

Adding \(20\%\) to \(\$68\) gives \(\$81.60\), which is wrong because that percentage would be calculated on the reduced price rather than the original. When the value after a change is given, always divide by the multiplier.

Simple interest is charged on the original amount borrowed for every year of the loan, so it grows in equal steps: \(I=P\times r\times t\) with \(r\) written as a decimal.

(a) \(9000\times0.08\times3\) [M1] oe. Take it in stages without a calculator: \(8\%\) of \(9000\) is \(720\), and \(3\times720=2160\). The interest is \(\$2160\) [A1].

(b) The total repaid is the amount borrowed plus the interest: \(9000+2160=\$11\,160\) [B1] ft.

(c) The \(36\) payments share the total repaid equally: \(11\,160\div36\) [M1] ft. Since \(36\times300=10\,800\) and the remaining \(360\) gives \(360\div36=10\), each payment is \(\$310\) [A1]. Note that \(36\) payments over \(3\) years is one payment a month, which is consistent.

(d) \(\frac{\text{interest}}{\text{amount borrowed}}=\frac{2160}{9000}\) [M1] ft. Divide both by \(360\): \(\frac{6}{25}\) [A1] cao. (Step by step: divide by \(10\) to get \(\frac{216}{900}\), then by \(9\) to get \(\frac{24}{100}\), then by \(4\) to get \(\frac{6}{25}\).)

Contrast this with compound interest, where the second year's interest would be charged on \(\$9720\) rather than on \(\$9000\). Simple interest keeps the base fixed, which is why the yearly charge here is \(\$720\) every time.

The angle between a space diagonal and the base is measured between the diagonal itself and its projection onto the base, which is the diagonal of the rectangular base. Finding it takes two stages.

1. Base diagonal \(=\sqrt{6^{2}+8^{2}}\) [M1] \(=\sqrt{36+64}=\sqrt{100}=10\) cm [A1], using the 6, 8, 10 right-angled triangle.
2. The space diagonal, the base diagonal and the vertical edge form a right-angled triangle, with the right angle where the vertical edge meets the base. The vertical edge of 10 cm is opposite the required angle \(\theta\), and the base diagonal of 10 cm is adjacent to it, so \(\tan\theta=\dfrac{10}{10}\) [M1].
3. \(\tan\theta=1\), so \(\theta=45^{\circ}\) [A1] cao.

The result is exact and needs no calculator, because a tangent of 1 corresponds to an isosceles right-angled triangle in which the opposite and adjacent sides are equal. That is exactly the situation here, since the height happens to equal the base diagonal.

The step most often mishandled is choosing which length is adjacent. It is not one of the base edges, 6 cm or 8 cm, but the base diagonal, because that is the line directly beneath the space diagonal.

Reading a Venn diagram means recognising which regions each piece of set notation covers. Here the diagram's four regions are the badminton-only region, the overlap of the two circles, the squash-only region and the outside.

(a) \(n(B)\) is the whole badminton circle, both its own region and the overlap it shares with squash: \(n(B)=20\) [B1]

(b) \((B\cup S)'\) is the region outside both circles, the members who play neither game: \(9\) [B1]

(c) \(B'\cap S\) means not badminton and squash, so it is the squash-only region: \(11\) [B1]

Since the four regions total \(40\) members, the overlap holds \(6\) and the badminton-only region holds \(20-6=14\).

(d) Playing both games is the overlap: \(P=\frac{6}{40}=\frac{3}{20}\) [B1] oe, dividing numerator and denominator by \(2\).

(e) Exactly one game means the two "only" regions, deliberately excluding the overlap:

\(P=\frac{14+11}{40}\) [M1] \(=\frac{25}{40}=\frac{5}{8}\) [A1] oe

Note the difference between \(n(B)=20\), which includes the members who also play squash, and the badminton-only count of \(14\). The word "exactly" is the signal to exclude the overlap.

The key geometry fact is the exterior angle theorem: the exterior angle of a triangle equals the sum of the two interior angles not adjacent to it. Here \(ACD\) is the exterior angle at \(C\), and the two opposite interior angles are \(BAC\) and \(ABC\).

(a) Show that \(x=30\)

1. Apply the theorem: \(4x-25=x+15+2x-10\) [M1] for equating the exterior angle to the sum of the two opposite interior angles.
2. Collect like terms on the right: \(x+2x=3x\) and \(15-10=+5\), giving \(4x-25=3x+5\) [M1].
3. Subtract \(3x\) from both sides: \(x-25=5\), then add 25 to both sides: \(x=30\) [A1].

An equally valid route uses angles on a straight line at \(C\): angle \(ACB=180-(4x-25)=205-4x\), and the three interior angles sum to \(180^{\circ}\). That gives the same value of \(x\).

(b) Substitute \(x=30\) into each expression.

The third angle is found either from the angle sum of the triangle, \(180-45-50=85\), or from the straight line \(BCD\), since the exterior angle is \(4(30)-25=95^{\circ}\) and \(180-95=85^{\circ}\). Both give \(85^{\circ}\), and the check \(45+50+85=180\) confirms the work.

The frequent error is treating \(ACD\) as if it were an interior angle and including it in the \(180^{\circ}\) sum. Only \(BAC\), \(ABC\) and \(ACB\) are interior to the triangle.

The circumference of a circle is \(C=2\pi r\), or equivalently \(\pi d\) using the diameter. With \(r=5\) cm,

\(C=2\times\pi\times 5=10\pi\) cm [B1] cao

Leaving the answer in terms of \(\pi\) means the symbol stays in the answer, so no decimal approximation is used and the value is exact.

The usual confusion is with the area formula \(A=\pi r^{2}\), which would give \(25\pi\). Circumference is a length, so it involves \(r\) to the first power and its unit is cm; area involves \(r^{2}\) and its unit would be cm\(^{2}\). Checking the unit of the answer is a quick way to confirm the right formula has been used.

This quadratic has a leading coefficient of 6, so factorising needs a pair of numbers that multiply to \(6\times(-10)=-60\) and add to \(+11\). Those numbers are \(+15\) and \(-4\).

1. Split the middle term: \(6x^{2}+15x-4x-10=0\).
2. Factorise in pairs: \(3x(2x+5)-2(2x+5)=0\).
3. Take out the common bracket: \((3x-2)(2x+5)=0\) [M1]. Correct use of the quadratic formula also earns this mark.

A product equals zero only when one of its factors is zero, so each bracket is set to zero in turn.

1. \(3x-2=0\) gives \(x=\dfrac{2}{3}\) [A1].
2. \(2x+5=0\) gives \(x=-\dfrac{5}{2}\) [A1] oe.

Check the factorisation by expanding: \(6x^{2}+15x-4x-10=6x^{2}+11x-10\), which matches the original equation.

The mistake to avoid is reading the roots straight off the brackets as 2 and \(-5\). Each bracket must actually be solved, because the coefficients of \(x\) are not 1.

(a) Each corner loses a quarter of a circle of radius 4 cm. Four quarters make one complete circle [M1], so the total area removed is the area of a single circle of radius 4 cm.

Area removed \(=\pi r^{2}=\pi\times 4^{2}=16\pi\) cm\(^{2}\) [A1].

(b) The whole plate has area \(20\times 12=240\) cm\(^{2}\) [M1], so the metal remaining is

\((240-16\pi)\) cm\(^{2}\) [A1].

The two terms cannot be combined, because one is a plain number and the other is a multiple of \(\pi\). Leaving the answer in this form keeps it exact.

(c) Taking \(\pi=3.14\): \(16\times 3.14=50.24\), so the remaining area is \(240-50.24=189.76\) cm\(^{2}\), which to the nearest square centimetre is 190 cm\(^{2}\) [B1] cao.

The idea worth carrying forward is that quarter circles cut from the four corners of a rectangle always combine into exactly one circle, provided they all have the same radius. That turns four separate calculations into one.

Compound interest is applied to the balance at the start of each year, so a rate of \(10\%\) multiplies the amount by \(1.1\) every year.

1. Value after 2 years \(=2000\times 1.1^{2}\) [M1] oe.
2. \(1.1^{2}=1.21\), so the value is \(2000\times 1.21=\$2420\) [M1].
3. The interest earned is the growth on top of the original investment: \(2420-2000=\$420\) [A1] cao.

Year by year the same result appears: \(10\%\) of \(\$2000\) is \(\$200\), giving \(\$2200\) after one year, and \(10\%\) of \(\$2200\) is \(\$220\), giving \(\$2420\). The second year earns \(\$20\) more than the first, and that \(\$20\) is the interest on the first year's interest, which is what compounding means.

Simple interest at the same rate would earn \(2\times\$200=\$400\), so the compound arrangement is \(\$20\) better. The final step matters: the question asks for the interest, not the total value, so \(\$2420\) on its own would not answer it.

Coordinate work with vectors uses three ideas: the connecting vector is finish minus start, the length comes from Pythagoras, and a point dividing a line in a given ratio is reached by travelling the matching fraction of the connecting vector.

(a) \(\overrightarrow{AB}=\binom{7-(-1)}{-4-4}=\binom{8}{-8}\) [B1]

(b) \(|\overrightarrow{AB}|=\sqrt{8^{2}+(-8)^{2}}=\sqrt{64+64}=\sqrt{128}\) [M1] ft. Take out the largest square factor: \(128=64\times2\), so \(\sqrt{128}=8\sqrt{2}\) [A1] cao, giving \(k=8\).

(c) The midpoint averages the coordinates: \(\left(\frac{-1+7}{2},\frac{4+(-4)}{2}\right)\) [M1] \(=(3,0)\) [A1]

(d) \(AP:PB=3:1\) divides \(AB\) into \(4\) equal parts, with \(P\) three parts along from \(A\):

1. \(\overrightarrow{AP}=\frac{3}{4}\overrightarrow{AB}=\frac{3}{4}\binom{8}{-8}=\binom{6}{-6}\) [M1]
2. Add this to the position of \(A\): \((-1+6,\;4-6)\) [M1] ft
3. \(P\) is \((5,-2)\) [A1]

The fraction is \(\frac{3}{3+1}=\frac{3}{4}\), not \(\frac{3}{1}\); the denominator is the total number of parts. A check is that \(P\) lies beyond the midpoint \((3,0)\) and closer to \(B(7,-4)\), which matches \(AP\) being three times \(PB\).

Clear both fractions by multiplying every term by the common denominator \((x+1)(x-1)\). This turns the equation into a quadratic.

1. Multiplying through gives \(4(x-1)+4(x+1)=3(x+1)(x-1)\) [M1].
2. The left side is \(4x-4+4x+4=8x\). The right side is a difference of two squares, \(3(x^{2}-1)=3x^{2}-3\), so \(8x=3x^{2}-3\) [M1].
3. Rearrange to the standard form: \(3x^{2}-8x-3=0\). Factorising needs two numbers multiplying to \(3\times(-3)=-9\) and adding to \(-8\), namely \(-9\) and \(+1\), which gives \((3x+1)(x-3)=0\) [M1] oe.
4. Setting each factor to zero gives \(x=3\) and \(x=-\dfrac{1}{3}\) [A1] oe.

Both values are valid, since neither makes a denominator zero; only \(x=1\) or \(x=-1\) would have to be rejected.

Check \(x=3\): \(\dfrac{4}{4}+\dfrac{4}{2}=1+2=3\), as required.

A vector \(\binom{X}{Y}\) is parallel to \(\binom{1}{-1}\) when it is a scalar multiple of it, which means \(Y=-X\): the vertical component must be the negative of the horizontal one.

1. Build the combination: \(\mathbf{p}+k\mathbf{q}=\binom{4}{-3}+k\binom{-2}{1}=\binom{4-2k}{-3+k}\) [M1]
2. Impose the parallel condition: \(-3+k=-(4-2k)\) [M1] oe
3. Expand the right-hand side: \(-3+k=-4+2k\) [M1]
4. Rearranging, \(-3+4=2k-k\), so \(k=1\) [A1]

Check the result: with \(k=1\), \(\mathbf{p}+\mathbf{q}=\binom{2}{-2}=2\binom{1}{-1}\), which is indeed parallel. The condition to state is that the components are in the ratio \(1:-1\); setting the two components equal to each other instead of opposite is the usual error, since \(\binom{1}{-1}\) points down and to the right.

The line of best fit summarises the trend in the data, so its equation lets one variable be predicted from the other. Substituting a known value into the equation is all that is required.

(a) Put \(x=6\) into \(y=-3x+40\):

\(y=-3(6)+40=-18+40=22\) [B1] cao

(b) Here \(y\) is known and \(x\) is wanted, so form an equation and solve it:

1. \(10=-3x+40\) [M1] oe
2. \(3x=40-10=30\)
3. \(x=10\) [A1] cao

The negative gradient of \(-3\) means \(y\) falls by \(3\) for every increase of \(1\) in \(x\), which is negative correlation. Be careful not to substitute \(10\) for \(x\) in part (b); the value given is a \(y\) value, and the two variables are not interchangeable.

Pattern \(n\) is an \(n\) by \(n\) grid of small squares, so the number of small squares is a square number, while the matchsticks form the horizontal and vertical lines of the grid.

(a) For Pattern \(4\): small squares \(=4^{2}=16\) [B1], and matchsticks \(=40\) [B1], continuing the pattern \(4,12,24,40\) whose differences are \(8,12,16\).

(b) An \(n\) by \(n\) grid holds \(n^{2}\) small squares [B1].

(c) The matchstick counts have first differences \(8,12,16\) and a constant second difference of \(4\), so the rule is quadratic with \(n^{2}\) coefficient \(\frac{4}{2}=2\), giving \(2n^{2}\) [M1]. Subtracting \(2n^{2}=2,8,18,32\) from \(4,12,24,40\) leaves \(2,4,6,8\), which is \(2n\) [M1]. Hence

Matchsticks \(=2n^{2}+2n\) [A1] oe, which factorises as \(2n(n+1)\)

The structure of the grid confirms this: there are \(n+1\) horizontal lines each made of \(n\) matches, and \(n+1\) vertical lines each made of \(n\) matches, giving \(2n(n+1)\).

(d) Set the expression equal to \(220\):

1. \(2n^{2}+2n=220\) [M1]
2. Divide by \(2\) and rearrange: \(n^{2}+n-110=0\), which factorises as \((n+11)(n-10)=0\) [M1] oe, since \(11\times(-10)=-110\) and \(11-10=1\).
3. The roots are \(n=-11\) and \(n=10\). A pattern number cannot be negative, so \(n=10\) [A1]

Check: \(2(100)+2(10)=200+20=220\). Rejecting the negative root by referring to the context is expected in pattern questions, since only positive whole numbers make sense as pattern numbers.

A value given to \(1\) decimal place lies within half of \(0.1\), that is \(0.05\), of the stated figure.

So \(8.55 \leqslant x \lt 8.65\) and \(3.15 \leqslant y \lt 3.25\) [M1].

For a subtraction, the result is smallest when the first number is as small as possible and the number being taken away is as large as possible. That means using the lower bound of \(x\) with the upper bound of \(y\):

\(8.55 - 3.25\) [M1] \(= 5.3\) [A1] cao.

Using \(8.55 - 3.15 = 5.4\) is the common error; that pairing gives neither bound of the difference. As a contrast, the upper bound of \(x - y\) would be \(8.65 - 3.15 = 5.5\), so the true value of \(x - y\) lies between \(5.3\) and \(5.5\).

Exam takeaway: for addition pair like with like, but for subtraction and division the bounds must be crossed over.

Two vectors are parallel exactly when one is a scalar multiple of the other, which means both components are multiplied by the same number.

1. Compare the top components to find the scale factor: \(6\div2=3\) [M1]
2. The same factor must apply to the bottom components: \(k=3\times(-5)=-15\) [A1]

The single scale factor is the whole point. Choosing \(k=-5\) treats the components as unrelated, and adding \(4\) to each component (since \(6=2+4\)) would be a translation of the vector, not a scalar multiple, so it does not preserve direction. A quick check is that \(\binom{6}{-15}=3\binom{2}{-5}\).

Bearings are measured clockwise from north, using three figures. The key fact is that the north lines at \(A\) and at \(B\) are parallel, so the bearing of \(A\) from \(B\) (the back bearing) is the original bearing plus \(180^{\circ}\).

(a) The ship sails from \(A\) to \(B\) on a bearing of \(060^{\circ}\), so the direction from \(B\) back to \(A\) is

\(060+180=240^{\circ}\) [M1]

Both \(240^{\circ}\) and the bearing \(150^{\circ}\) of \(C\) from \(B\) are measured clockwise from the same north line at \(B\), so the angle between \(BA\) and \(BC\) is the difference:

\(240-150=90\), so angle \(ABC=90^{\circ}\) [A1]

Subtracting the two original bearings, \(150-60=90\), gives the same number by coincidence of the figures but is not a valid argument, because those two angles are measured at different points. The back bearing step is what makes the reasoning correct.

(b) Triangle \(ABC\) is right-angled at \(B\) with legs \(AB=8\) km and \(BC=6\) km, so Pythagoras' theorem applies with \(AC\) as the hypotenuse:

\(AC^{2}=8^{2}+6^{2}=64+36=100\) [M1]

\(AC=\sqrt{100}=10\) km [A1] cao

The \(6\), \(8\), \(10\) triangle is a scaled \(3\), \(4\), \(5\) triangle, which is why the arithmetic works out exactly without a calculator.

The two days are independent, so the probability of a particular pair of outcomes is the product of the separate probabilities.

First find the probability of no rain on one day. Raining and not raining are the only two possibilities, so they sum to \(1\):

\(P(\text{no rain}) = 1 - 0.35 = 0.65\).

For no rain on both days, multiply: \(0.65 \times 0.65\) [M1].

By hand, \(65 \times 65 = 4225\), and there are four decimal places in total, so the answer is \(0.4225\) oe [A1]. As a fraction this is \(\dfrac{169}{400}\).

Adding the probabilities would give \(1.3\), which is impossible since no probability can exceed \(1\). That check alone rules out addition. Note also that the answer is smaller than \(0.65\), as it must be: requiring two events together is harder than requiring one.

Both values come from the half of an equilateral triangle, the right-angled triangle with sides 1, \(\sqrt{3}\) and 2.

1. \(\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}\), the side opposite \(60^{\circ}\) over the hypotenuse.
2. \(\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}\), the side adjacent to \(30^{\circ}\) over the hypotenuse.

These are equal because the two angles are complementary: the side opposite one is the side adjacent to the other. So the product is

\(\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{3}}{2}\) [M1].

Multiply numerators and denominators: \(\sqrt{3}\times\sqrt{3}=3\) and \(2\times 2=4\), giving

\(\dfrac{3}{4}\) [A1] cao.

The step worth fixing is \(\sqrt{3}\times\sqrt{3}=3\). Squaring a square root returns the original number, so the surd disappears and the answer is a plain fraction. Writing \(\sqrt{9}\) or \(\sqrt{6}\) here are the two common errors.

Factorising by grouping works when four terms split into two pairs that leave the same bracket behind. Take the first two terms together and the last two together.

1. From \(2ax-6ay\) the common factor is \(2a\), giving \(2a(x-3y)\) [M1].
2. From \(bx-3by\) the common factor is \(b\), giving \(b(x-3y)\) [M1].
3. Both pairs now contain the bracket \((x-3y)\), so that bracket is itself a common factor: \(2a(x-3y)+b(x-3y)=(2a+b)(x-3y)\) [A1] oe.

Check by expanding: \((2a+b)(x-3y)=2ax-6ay+bx-3by\), which is the original expression.

The signal that the grouping is right is that both brackets match exactly. If they come out as \((x-3y)\) and \((3y-x)\), the second pair has been factorised with the wrong sign; take out \(-b\) instead of \(b\) to make them agree.

The condition \(20 \lt n \lt 40\) is strict at both ends, so the possible values run from \(21\) to \(39\).

(a) A prime number has exactly two factors, itself and \(1\). Testing each candidate for divisibility by \(2\), \(3\) and \(5\) removes almost all of them, and it is enough to test primes up to \(6\), since \(6^{2} = 36\) is near the top of the range and \(7^{2} = 49\) is above it.

The primes in the range are \(23,\ 29,\ 31,\ 37\). Three of these correct with no incorrect extras earns [B1], and the complete list earns the second [B1].

Numbers such as \(21 = 3 \times 7\), \(27 = 3 \times 9\), \(33 = 3 \times 11\) and \(39 = 3 \times 13\) are the ones most often included by mistake.

(b) The cube numbers near this range are \(2^{3} = 8\), \(3^{3} = 27\) and \(4^{3} = 64\). Only \(27\) lies between \(20\) and \(40\), so \(n = 27\) [B1].

(c) The square numbers are \(4^{2} = 16\), \(5^{2} = 25\), \(6^{2} = 36\) and \(7^{2} = 49\). Those in range are \(25\) and \(36\) [B1].

(a) The volume of a cylinder is \(V = \pi r^{2} h\). Substituting the known values gives an equation in \(r\):

\(\pi r^{2} \times 20 = 500\pi\). Dividing both sides by \(\pi\) and then by \(20\) gives \(r^{2} = \dfrac{500}{20} = 25\) [M1].

\(r = \sqrt{25} = 5\) cm [A1] cao. The negative root is rejected because a radius is a length.

(b) The curved surface of a cylinder unrolls into a rectangle whose width is the circumference \(2\pi r\) and whose height is \(h\), so its area is \(2\pi r h\).

\(2\pi \times 5 \times 20\) [M1] \(= 200\pi\) cm\(^{2}\) [A1] cao.

Cancelling \(\pi\) at the start of part (a) is what makes the arithmetic exact without a calculator. Note that the question asks only for the curved surface; the total surface area of this closed cylinder would additionally include two circular ends of \(25\pi\) cm\(^{2}\) each.

Two sides and the angle between them are given, which is exactly the case for the cosine rule. The angle \(120^{\circ}\) is obtuse, so its cosine is negative, and the exact values \(\cos 120^{\circ}=-\dfrac{1}{2}\) and \(\sin 120^{\circ}=\dfrac{\sqrt{3}}{2}\) make the whole question possible without a calculator.

(a) The side \(PR\) is opposite the given angle \(PQR\), so

\(PR^{2}=6^{2}+10^{2}-2\times 6\times 10\times\cos 120^{\circ}\) [M1]

Using \(\cos 120^{\circ}=-\dfrac{1}{2}\) [M1], the final term becomes \(-120\times\left(-\dfrac{1}{2}\right)=+60\), so

\(PR^{2}=36+100+60=196\), hence \(PR=\sqrt{196}=14\) cm [A1]

The sign is the whole point here: because the angle is obtuse, the subtraction turns into an addition and \(PR\) comes out longer than either given side, as it must be when the angle between them is wide.

(b) Area of a triangle from two sides and the included angle is \(\dfrac{1}{2}ab\sin C\):

\(\dfrac{1}{2}\times 6\times 10\times\dfrac{\sqrt{3}}{2}\) [M1]

\(\dfrac{1}{2}\times 6\times 10=30\), and \(30\times\dfrac{\sqrt{3}}{2}=15\sqrt{3}\).

Area \(=15\sqrt{3}\) cm\(^{2}\) [A1]

"Exact" means the surd must be left in place; a rounded decimal would not earn the mark.

(c) Apply the sine rule, pairing each angle with the side opposite it. Angle \(QPR\) is opposite \(QR=10\), and angle \(PQR=120^{\circ}\) is opposite \(PR=14\):

\(\dfrac{\sin QPR}{10}=\dfrac{\sin 120^{\circ}}{14}\), so \(\sin QPR=\dfrac{10}{14}\times\dfrac{\sqrt{3}}{2}=\dfrac{10\sqrt{3}}{28}=\dfrac{5\sqrt{3}}{14}\)

\(\sin QPR=\dfrac{5\sqrt{3}}{14}\) oe [B1]