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Angles, polygons, congruence and vectors — solutions

16 full solutions · spec M5.1-M5.6, M5.10, M5.14, M5.18, M5.19

Section 1 Part 2: Higher GCSE mathematics16 questions
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Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. GEO-01Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★☆☆

The interior angle of a regular polygon is \(156^\circ\). How many sides does it have?

  1. A\(24\)
  2. B\(20\)
  3. C\(10\)
  4. D\(12\)
  5. E\(18\)
  6. F\(15\)
Answer: F. Exterior angle \(24^\circ\); \(360/24 = 15\) sides.
Full method — how and why
Work with the exterior angle: they always sum to \(360^\circ\), which makes the arithmetic trivial.
  1. Exterior angle \(= 180^\circ - 156^\circ = 24^\circ\).
    Interior and exterior angles at a vertex lie on a straight line.
  2. The exterior angles of any polygon sum to \(360^\circ\), and for a regular polygon they are all equal.
    Walking once around the shape turns you through a full circle.
  3. Number of sides \(= \dfrac{360}{24} = 15\).
    Divide the total turn by the turn per vertex.
Where it goes wrong
  • Using \(\frac{360}{156}\) or \(\frac{156}{24}\).
  • Using the interior-angle-sum formula and then having to solve \(\frac{180(n-2)}{n} = 156\) - correct but slower.
Takeaway For regular polygons, go via the exterior angle: \(n = \frac{360}{\text{exterior}}\).
2. GEO-02Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★☆☆

A polygon has interior angles that sum to \(1980^\circ\). How many sides does it have?

  1. A\(10\)
  2. B\(15\)
  3. C\(14\)
  4. D\(11\)
  5. E\(13\)
  6. F\(12\)
Answer: E. \(180(n - 2) = 1980 \Rightarrow n - 2 = 11 \Rightarrow n = 13\).
Full method — how and why
Direct application of the interior-angle-sum formula.
  1. Recall the formula: the interior angles of an \(n\)-gon sum to \(180(n-2)^\circ\).
    The polygon splits into \(n - 2\) triangles, each contributing \(180^\circ\).
  2. Set up: \(180(n - 2) = 1980\).
    Substitute the given total.
  3. Divide: \(n - 2 = 11\), so \(n = 13\).
    \(1980 \div 180 = 11\); remember to add the \(2\) back.
Where it goes wrong
  • Forgetting to add \(2\) and answering \(11\) - an offered option.
  • Using \(180n\) instead of \(180(n-2)\).
Takeaway Interior sum \(= 180(n-2)\): divide, then add \(2\).
3. GEO-03Angles, polygons, congruence and vectors · spec M5.19 · Paper 1 style · ★☆☆

\(\mathbf{a} = \binom{2}{-1}\) and \(\mathbf{b} = \binom{-3}{4}\). Find \(2\mathbf{a} - \mathbf{b}\).

  1. A\(\binom{5}{-5}\)
  2. B\(\binom{-7}{6}\)
  3. C\(\binom{-1}{3}\)
  4. D\(\binom{1}{2}\)
  5. E\(\binom{7}{-2}\)
  6. F\(\binom{7}{-6}\)
Answer: F. \(\binom{4}{-2} - \binom{-3}{4} = \binom{7}{-6}\).
Full method — how and why
Column-vector arithmetic: operate on each component separately.
  1. Double \(\mathbf a\): \(2\binom{2}{-1} = \binom{4}{-2}\).
    A scalar multiplies both components.
  2. Subtract \(\mathbf b\) component by component: \(\binom{4 - (-3)}{-2 - 4}\).
    Subtracting a negative adds - the top component becomes \(4 + 3\).
  3. Simplify: \(\binom{7}{-6}\).
    Both components computed independently.
Where it goes wrong
  • Adding \(\mathbf b\) instead of subtracting, giving \(\binom{1}{2}\) - an offered option.
  • Getting the sign of the bottom component wrong: \(-2 - 4 = -6\), not \(+2\).
Takeaway Vectors combine component by component; watch double negatives.
4. GEO-05Angles, polygons, congruence and vectors · spec M5.5 · Paper 1 style · ★☆☆

Two similar triangles have corresponding sides of length \(6\) cm and \(9\) cm. The smaller triangle has area \(20\) cm\(^2\). Find the area of the larger triangle.

  1. A\(30\) cm\(^2\)
  2. B\(67.5\) cm\(^2\)
  3. C\(40\) cm\(^2\)
  4. D\(45\) cm\(^2\)
  5. E\(27\) cm\(^2\)
  6. F\(60\) cm\(^2\)
Answer: D. Area scale factor \(\left(\frac96\right)^2 = \frac94\): \(20 \times \frac94 = 45\).
Full method — how and why
Similar shapes: areas scale as the square of the length ratio.
  1. Length scale factor: \(\dfrac96 = \dfrac32\).
    From the smaller to the larger, so the factor is greater than \(1\).
  2. Area scale factor: \(\left(\dfrac32\right)^2 = \dfrac94\).
    Both dimensions scale, so the area scales by the square.
  3. Multiply: \(20 \times \dfrac94 = 45\) cm\(^2\).
    \(20 \div 4 = 5\), then \(\times 9\).
Where it goes wrong
  • Using the length factor directly: \(20 \times \frac32 = 30\) - an offered option.
  • Cubing (that would be for volumes), giving \(67.5\).
Takeaway Area ratio \(=\) (length ratio)\(^2\); check the direction of the scale factor.
5. GEO-07Angles, polygons, congruence and vectors · spec M5.6 · Paper 1 style · ★☆☆

The triangle with vertices \((1, 1)\), \((3, 1)\), \((1, 4)\) is enlarged with scale factor \(-2\), centre the origin. Which of the following is a vertex of the image?

  1. A\((-1, -4)\)
  2. B\((6, 2)\)
  3. C\((-2, -8)\)
  4. D\((-6, -8)\)
  5. E\((-3, -1)\)
  6. F\((2, 8)\)
Answer: C. Multiply each coordinate by \(-2\): \((-2, -2)\), \((-6, -2)\), \((-2, -8)\).
Full method — how and why
Enlargement with a negative scale factor about the origin: multiply every coordinate by that factor.
  1. With centre the origin, enlargement by \(k\) sends \((x,y)\) to \((kx, ky)\).
    Position vectors scale directly when the centre is the origin.
  2. Apply \(k = -2\) to each vertex: \((1,1) \to (-2,-2)\); \((3,1) \to (-6,-2)\); \((1,4) \to (-2,-8)\).
    A negative factor also rotates the shape through \(180^\circ\) - the image is on the opposite side of the origin.
  3. Check the options: \((-2,-8)\) appears.
    Match against the computed image vertices.
Where it goes wrong
  • Using \(+2\) and choosing \((2,8)\) - an offered option.
  • Multiplying only one coordinate.
Takeaway Enlargement about the origin: multiply both coordinates by \(k\); a negative \(k\) flips the shape through the centre.
6. GEO-08Angles, polygons, congruence and vectors · spec M5.4 · Paper 1 style · ★☆☆

Which of the following sets of information is NOT sufficient to prove that two triangles are congruent?

  1. Atwo sides and a non-included angle equal (SSA)
  2. Btwo sides and the included angle equal (SAS)
  3. Ctwo angles and a corresponding side equal (ASA)
  4. Dright angle, hypotenuse and one other side equal (RHS)
  5. Ethree sides equal (SSS)
Answer: A. SSA is not a congruence condition (it is the ambiguous case of the sine rule); the other four are the standard criteria.
Full method — how and why
The four congruence criteria are SSS, SAS, ASA and RHS. Anything else - notably SSA - is not sufficient.
  1. Recall the valid criteria: SSS, SAS, ASA (or AAS), RHS.
    These four are the ones that fix a triangle completely.
  2. Consider SSA: two sides and a non-included angle. This is the ambiguous case of the sine rule, where two different triangles can share the same data.
    The side opposite the known angle can 'swing' to two positions - exactly the situation in TRI-07.
  3. So SSA is the odd one out.
    RHS is really a special case of SSA that works, because the right angle removes the ambiguity.
Where it goes wrong
  • Thinking RHS is invalid because it looks like SSA - the right angle makes it unambiguous.
  • Confusing AAA (similar but not congruent) with ASA.
Takeaway SSS, SAS, ASA, RHS prove congruence; SSA does not, and AAA gives only similarity.
7. GEO-09Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★☆☆

In the diagram (not shown), \(AB\) is parallel to \(CD\). A transversal makes an angle of \((3x + 10)^\circ\) with \(AB\) and the co-interior angle with \(CD\) is \((2x - 5)^\circ\). Find \(x\).

  1. A\(25\)
  2. B\(55\)
  3. C\(15\)
  4. D\(30\)
  5. E\(35\)
  6. F\(45\)
Answer: E. Co-interior angles sum to \(180^\circ\): \(5x + 5 = 180 \Rightarrow x = 35\).
Full method — how and why
Co-interior (allied) angles between parallel lines add to \(180^\circ\).
  1. Identify the relationship: co-interior angles lie on the same side of the transversal, between the parallel lines.
    Their names matter: alternate angles are equal, corresponding angles are equal, co-interior angles are supplementary.
  2. Write the equation: \((3x + 10) + (2x - 5) = 180\).
    Supplementary means summing to \(180^\circ\).
  3. Simplify: \(5x + 5 = 180\), so \(5x = 175\) and \(x = 35\).
    Combine like terms, then divide.
Where it goes wrong
  • Setting the angles equal (treating them as alternate), which gives \(x = 15\) - an offered option.
  • Arithmetic: \(10 - 5 = 5\), not \(15\).
Takeaway With parallel lines: alternate equal, corresponding equal, co-interior sum to \(180^\circ\).
8. GEO-10Angles, polygons, congruence and vectors · spec M5.3 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IEvery rhombus is a parallelogram.
  2. IIEvery rectangle is a rhombus.
  3. IIIThe diagonals of a kite are perpendicular.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: F. A rhombus has two pairs of parallel sides (I). A rectangle need not have equal sides (II false). The diagonals of a kite cross at right angles (III).
Full method — how and why
Quadrilateral classification. Each claim is about whether one family is contained in another.
  1. I: a rhombus has two pairs of parallel sides, which is the definition of a parallelogram. True.
    Rhombus \(=\) parallelogram with all sides equal, so it is a special parallelogram.
  2. II: a rectangle has equal angles but not necessarily equal sides - a \(2 \times 5\) rectangle is not a rhombus. False.
    Only the square lies in both families.
  3. III: a kite's diagonals meet at right angles (the axis of symmetry bisects the other diagonal perpendicularly). True.
    This is a standard kite property.
  4. I and III only.
    Match to the option list.
Where it goes wrong
  • Reversing II: every rhombus is not a rectangle either, but the claim as stated is about rectangles.
  • Thinking a kite's diagonals bisect each other - only one bisects the other.
Takeaway Square \(=\) rectangle \(\cap\) rhombus; kites have perpendicular diagonals but only one is bisected.
9. GEO-12Angles, polygons, congruence and vectors · spec M5.6 · Paper 1 style · ★☆☆

A shape is reflected in the line \(y = x\) and then rotated \(90^\circ\) anticlockwise about the origin. The point \((3, 1)\) ends up at

  1. A\((1, 3)\)
  2. B\((-3, 1)\)
  3. C\((-1, 3)\)
  4. D\((1, -3)\)
  5. E\((3, -1)\)
  6. F\((-1, -3)\)
Answer: B. Reflection in \(y = x\): \((3, 1) \to (1, 3)\). Rotation \(90^\circ\) anticlockwise: \((x, y) \to (-y, x)\), so \((1, 3) \to (-3, 1)\).
Full method — how and why
Two transformations in sequence. Apply each rule in the stated order to the coordinates.
  1. Reflection in \(y = x\) swaps the coordinates: \((3,1) \to (1,3)\).
    The line \(y = x\) exchanges the roles of the two axes.
  2. Rotation \(90^\circ\) anticlockwise about the origin sends \((x,y) \to (-y,x)\).
    Worth memorising; check it on \((1,0) \to (0,1)\) ✓.
  3. Apply it: \((1,3) \to (-3,1)\).
    Here \(x = 1\), \(y = 3\), so the image is \((-3, 1)\).
Where it goes wrong
  • Applying the transformations in the wrong order.
  • Using the clockwise rule \((x,y) \to (y,-x)\), giving \((3,-1)\) - an offered option.
Takeaway Anticlockwise \(90^\circ\): \((x,y) \to (-y,x)\). Clockwise: \((x,y) \to (y,-x)\). Order matters.
10. GEO-13Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★☆☆

In triangle \(ABC\), angle \(A\) is twice angle \(B\), and angle \(C\) is \(20^\circ\) more than angle \(A\). Find angle \(B\).

  1. A\(64^\circ\)
  2. B\(30^\circ\)
  3. C\(32^\circ\)
  4. D\(36^\circ\)
  5. E\(40^\circ\)
  6. F\(28^\circ\)
Answer: C. \(B + 2B + (2B + 20) = 180 \Rightarrow 5B = 160 \Rightarrow B = 32^\circ\).
Full method — how and why
Express all three angles in terms of one, then use the angle sum of a triangle.
  1. Let angle \(B = b\). Then \(A = 2b\) and \(C = A + 20 = 2b + 20\).
    Express everything in terms of the angle being asked for.
  2. Angle sum: \(b + 2b + (2b + 20) = 180\).
    The three angles of a triangle sum to \(180^\circ\).
  3. Simplify: \(5b + 20 = 180\), so \(5b = 160\) and \(b = 32^\circ\).
    Check: \(32 + 64 + 84 = 180\) ✓.
Where it goes wrong
  • Setting \(C = 2b + 20\) but forgetting that \(C\) is \(20\) more than \(A\), not than \(B\).
  • Arithmetic: \(180 - 20 = 160\), not \(200\).
Takeaway Name one angle, express the rest in terms of it, then use the \(180^\circ\) sum.
11. GEO-14Angles, polygons, congruence and vectors · spec M5.5 · Paper 1 style · ★☆☆

In triangle \(ABC\), \(D\) lies on \(AB\) and \(E\) lies on \(AC\) with \(DE\) parallel to \(BC\). \(AD = 4\), \(DB = 6\) and \(DE = 5\). Find \(BC\).

  1. A\(12.5\)
  2. B\(8\)
  3. C\(15\)
  4. D\(10\)
  5. E\(7.5\)
  6. F\(12\)
Answer: A. Triangles \(ADE\) and \(ABC\) are similar with ratio \(AD : AB = 4 : 10\). \(BC = 5 \times \frac{10}{4} = 12.5\).
Full method — how and why
A line parallel to one side creates a similar triangle. Use the full side, not just the part.
  1. \(DE \parallel BC\) makes triangles \(ADE\) and \(ABC\) similar (equal corresponding angles).
    Parallel lines give equal corresponding angles, which is the AAA similarity condition.
  2. Compute the ratio: \(AB = AD + DB = 4 + 6 = 10\), so \(AD : AB = 4 : 10 = 2 : 5\).
    The ratio must compare corresponding whole sides - \(AD\) to \(AB\), not \(AD\) to \(DB\).
  3. Scale up \(DE\): \(BC = 5 \times \dfrac{10}{4} = 12.5\).
    Multiply by the reciprocal of the ratio to go from small to large.
Where it goes wrong
  • Using the ratio \(4 : 6\) and answering \(7.5\) - an offered option.
  • Dividing instead of multiplying, giving \(2\).
Takeaway In similar triangles compare corresponding full sides: \(\frac{AD}{AB}\), not \(\frac{AD}{DB}\).
12. X2-20Angles, polygons, congruence and vectors · spec M5.3 · Paper 2 style · ★☆☆

\(ABCD\) is a quadrilateral. \(P\): 'the diagonals of \(ABCD\) bisect each other'. \(Q\): '\(ABCD\) is a rectangle'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Answer: B. Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other: \(P\) is necessary. A non-rectangular parallelogram has bisecting diagonals without being a rectangle: not sufficient.
Full method — how and why
A quadrilateral hierarchy question. Bisecting diagonals characterise parallelograms, a larger class than rectangles.
  1. Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other. So \(Q \Rightarrow P\): \(P\) is necessary.
    Opposite sides equal and parallel forces the diagonals to share a midpoint.
  2. Is \(P\) sufficient? Take a non-rectangular parallelogram, e.g. a rhombus with angles \(60^\circ/120^\circ\): diagonals bisect each other, but it is not a rectangle.
    Bisecting diagonals say nothing about the angles.
  3. So \(P\) is necessary but not sufficient.
    For a rectangle you also need the diagonals to be equal in length.
Where it goes wrong
  • Answering 'necessary and sufficient' by confusing 'bisect each other' with 'bisect each other and are equal'.
  • Getting the direction backwards: the broader property (parallelogram) is necessary for the narrower one (rectangle).
Takeaway Diagonals bisect each other \(\iff\) parallelogram; add equal diagonals for a rectangle, add perpendicular for a rhombus.
13. GEO-04Angles, polygons, congruence and vectors · spec M5.19 · Paper 1 style · ★★☆

In triangle \(OAB\), \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\). \(M\) is the midpoint of \(AB\) and \(P\) is the point on \(OM\) with \(OP : PM = 2 : 1\). Express \(\overrightarrow{AP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

  1. A\(\dfrac{2}{3}(\mathbf{a} + \mathbf{b})\)
  2. B\(\dfrac{2}{3}\mathbf{b} - \dfrac{1}{3}\mathbf{a}\)
  3. C\(\dfrac{1}{3}(\mathbf{b} - \mathbf{a})\)
  4. D\(\dfrac{1}{3}\mathbf{b} - \dfrac{2}{3}\mathbf{a}\)
  5. E\(\dfrac{1}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}\)
  6. F\(\dfrac{1}{2}\mathbf{b} - \dfrac{1}{2}\mathbf{a}\)
Answer: D. \(\overrightarrow{OM} = \frac12(\mathbf a + \mathbf b)\), \(\overrightarrow{OP} = \frac23\overrightarrow{OM} = \frac13(\mathbf a + \mathbf b)\). \(\overrightarrow{AP} = \overrightarrow{OP} - \mathbf a = \frac13\mathbf b - \frac23\mathbf a\).
Full method — how and why
Vector geometry: build a route from \(A\) to \(P\) out of the known vectors, using the ratio to fix the fraction.
  1. Find \(\overrightarrow{OM}\): \(M\) is the midpoint of \(AB\), so \(\overrightarrow{OM} = \frac12(\mathbf a + \mathbf b)\).
    The midpoint of \(AB\) is the average of the two position vectors.
  2. Use the ratio \(OP : PM = 2 : 1\): \(P\) is \(\frac23\) of the way from \(O\) to \(M\), so \(\overrightarrow{OP} = \frac23 \times \frac12(\mathbf a + \mathbf b) = \frac13(\mathbf a + \mathbf b)\).
    \(2 : 1\) means \(2\) parts out of \(3\) - convert the ratio into a fraction of the whole.
  3. Travel from \(A\) to \(P\) via \(O\): \(\overrightarrow{AP} = \overrightarrow{AO} + \overrightarrow{OP} = -\mathbf a + \frac13(\mathbf a + \mathbf b)\).
    \(\overrightarrow{AO} = -\overrightarrow{OA} = -\mathbf a\); reversing a vector negates it.
  4. Simplify: \(-\mathbf a + \frac13\mathbf a + \frac13\mathbf b = \frac13\mathbf b - \frac23\mathbf a\).
    Collect the \(\mathbf a\) terms: \(-1 + \frac13 = -\frac23\).
Where it goes wrong
  • Using \(\frac12\) instead of \(\frac23\) for the ratio point.
  • Forgetting the \(-\mathbf a\) and giving \(\overrightarrow{OP}\) instead of \(\overrightarrow{AP}\) - an offered option.
Takeaway Route vectors: \(\overrightarrow{XY} = \overrightarrow{XO} + \overrightarrow{OY}\), and a ratio \(m : n\) means the fraction \(\frac{m}{m+n}\).
14. GEO-06Angles, polygons, congruence and vectors · spec M5.14 · Paper 1 style · ★★☆

A ship sails from \(P\) on a bearing of \(060^\circ\) for \(10\) km to \(Q\), then on a bearing of \(150^\circ\) for \(10\) km to \(R\). What is the bearing of \(P\) from \(R\)?

  1. A\(255^\circ\)
  2. B\(195^\circ\)
  3. C\(315^\circ\)
  4. D\(285^\circ\)
  5. E\(075^\circ\)
  6. F\(105^\circ\)
Answer: D. The turn at \(Q\) is \(90^\circ\) and \(PQ = QR\), so triangle \(PQR\) is right-angled isosceles and \(PR\) makes \(45^\circ\) with \(QR\). Bearing of \(R\) from \(P\) is \(060 + 45 = 105^\circ\); the back bearing is \(105 + 180 = 285^\circ\).
Full method — how and why
A bearings problem with a right angle. Find the shape of the triangle, then work out the bearing and reverse it.
  1. The bearing changes from \(060^\circ\) to \(150^\circ\), a turn of \(90^\circ\), so the path turns through a right angle at \(Q\).
    The interior angle \(PQR\) is therefore \(90^\circ\).
  2. Both legs are \(10\) km, so triangle \(PQR\) is right-angled isosceles, with the two base angles \(45^\circ\) each.
    Equal legs give equal base angles, each \(\frac{180 - 90}{2} = 45^\circ\).
  3. Bearing of \(R\) from \(P\): the direction \(PQ\) is \(060^\circ\), and \(PR\) is \(45^\circ\) further clockwise, so \(060 + 45 = 105^\circ\).
    \(R\) lies on the far side of \(PQ\) from north, so add.
  4. Reverse for the bearing of \(P\) from \(R\): add \(180^\circ\) to get \(285^\circ\).
    Back bearings differ by \(180^\circ\) (add if under \(180^\circ\), subtract if over).
Where it goes wrong
  • Giving \(105^\circ\), which is the bearing of \(R\) from \(P\) - the reverse of what is asked (and an offered option).
  • Subtracting \(180^\circ\) and getting a negative, or forgetting three-figure form.
Takeaway Bearings are clockwise from north; a back bearing differs by exactly \(180^\circ\).
15. GEO-11Angles, polygons, congruence and vectors · spec M5.19 · Paper 1 style · ★★☆

\(\overrightarrow{AB} = 3\mathbf{p} + 2\mathbf{q}\) and \(\overrightarrow{BC} = k\mathbf{p} - 4\mathbf{q}\), where \(\mathbf{p}\) and \(\mathbf{q}\) are not parallel. Given that \(A\), \(B\) and \(C\) are collinear, find \(k\).

  1. A\(2\)
  2. B\(6\)
  3. C\(-3\)
  4. D\(\dfrac{3}{2}\)
  5. E\(-6\)
  6. F\(-4\)
Answer: E. Collinear means \(\overrightarrow{BC} = \lambda\overrightarrow{AB}\): \(-4 = 2\lambda \Rightarrow \lambda = -2\), so \(k = 3\lambda = -6\).
Full method — how and why
Collinear points mean one vector is a scalar multiple of the other. Compare coefficients of the two independent directions.
  1. Collinear \(A\), \(B\), \(C\) means \(\overrightarrow{BC} = \lambda\overrightarrow{AB}\) for some scalar \(\lambda\).
    They share the point \(B\), so the two vectors must point along the same line.
  2. Write it out: \(k\mathbf p - 4\mathbf q = \lambda(3\mathbf p + 2\mathbf q)\).
    Because \(\mathbf p\) and \(\mathbf q\) are not parallel, coefficients can be compared independently.
  3. Compare \(\mathbf q\): \(-4 = 2\lambda\), so \(\lambda = -2\).
    Use the equation that contains no unknown besides \(\lambda\).
  4. Compare \(\mathbf p\): \(k = 3\lambda = -6\).
    Substitute the value of \(\lambda\).
Where it goes wrong
  • Taking \(\lambda = 2\) and answering \(6\) - an offered option.
  • Comparing \(\mathbf p\) with \(\mathbf q\) coefficients.
Takeaway Non-parallel vectors are independent: equate their coefficients separately to solve for scalars.
16. GEO-15Angles, polygons, congruence and vectors · spec M5.2 · Paper 1 style · ★★☆

A regular hexagon and a regular octagon share a common side. What is the size of the angle between their two other sides that meet at one end of the common side (the angle outside both polygons)?

  1. A\(120^\circ\)
  2. B\(95^\circ\)
  3. C\(105^\circ\)
  4. D\(135^\circ\)
  5. E\(115^\circ\)
  6. F\(75^\circ\)
Answer: C. Angles at the shared vertex: \(120^\circ\) (hexagon) \(+ 135^\circ\) (octagon) \(+ x = 360^\circ\), so \(x = 105^\circ\).
Full method — how and why
Angles at a point sum to \(360^\circ\). Find each polygon's interior angle and subtract.
  1. Hexagon interior angle: exterior \(= \frac{360}{6} = 60^\circ\), so interior \(= 120^\circ\).
    Regular polygon: interior \(= 180 - \frac{360}{n}\).
  2. Octagon interior angle: exterior \(= \frac{360}{8} = 45^\circ\), so interior \(= 135^\circ\).
    Same formula with \(n = 8\).
  3. At the shared vertex three angles meet and fill a full turn: \(120 + 135 + x = 360\).
    The two interior angles plus the gap between the outer sides make a complete revolution.
  4. Solve: \(x = 360 - 255 = 105^\circ\).
    That gap is the angle asked for.
Where it goes wrong
  • Using \(180^\circ\) instead of \(360^\circ\) at the vertex.
  • Using exterior angles in the sum, giving \(255^\circ\) or similar.
Takeaway Regular \(n\)-gon interior angle \(= 180 - \frac{360}{n}\); angles round a point total \(360^\circ\).