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Triangles, radians and sectors — solutions

17 full solutions · spec MM4.1, MM4.2, M5.16

Section 1 Part 1: AS-level pure mathematics17 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. TRI-01Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★☆☆

In triangle \(ABC\), \(AB = 5\), \(AC = 8\) and angle \(BAC = 60^\circ\). Find \(BC\).

  1. A\(\sqrt{129}\)
  2. B\(\sqrt{49 + 40\sqrt3}\)
  3. C\(\sqrt{89}\)
  4. D\(\sqrt{69}\)
  5. E\(9\)
  6. F\(7\)
Answer: F. Cosine rule: \(BC^2 = 25 + 64 - 2\cdot5\cdot8\cdot\frac12 = 89 - 40 = 49\), so \(BC = 7\).
Full method — how and why
Two sides and the angle between them: that is exactly the cosine rule's input.
  1. Choose the rule: you know two sides and the included angle, so use the cosine rule, not the sine rule.
    The sine rule needs a side opposite a known angle; here the known angle sits between the known sides.
  2. Write it with \(BC\) opposite the known angle \(A\): \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A\).
    The side being found must be opposite the angle being used.
  3. Substitute, using \(\cos 60^\circ = \frac12\): \(BC^2 = 25 + 64 - 2(5)(8)\left(\frac12\right) = 89 - 40 = 49\).
    \(\cos 60^\circ = \frac12\) is one of the exact values you must recall.
  4. Take the square root: \(BC = 7\).
    Lengths are positive, so only the positive root.
Where it goes wrong
  • Forgetting the \(-2bc\cos A\) term and answering \(\sqrt{89}\) - which is offered.
  • Using \(\cos 60^\circ = \frac{\sqrt3}{2}\) (that is \(\cos 30^\circ\)), giving \(\sqrt{89 - 40\sqrt3}\).
Takeaway Two sides plus the included angle \(\Rightarrow\) cosine rule; the unknown side is opposite the known angle.
2. TRI-02Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★☆☆

A triangle has sides \(4\), \(5\) and \(6\). Find the cosine of its largest angle.

  1. A\(\dfrac{9}{16}\)
  2. B\(\dfrac{1}{8}\)
  3. C\(\dfrac{5}{8}\)
  4. D\(-\dfrac{1}{8}\)
  5. E\(\dfrac{1}{4}\)
  6. F\(\dfrac{3}{4}\)
Answer: B. The largest angle is opposite the side \(6\): \(\cos\theta = \dfrac{16 + 25 - 36}{2\cdot4\cdot5} = \dfrac{5}{40} = \dfrac18\).
Full method — how and why
Three sides given, an angle wanted: the rearranged cosine rule. The word 'largest' tells you which angle.
  1. Identify the largest angle: it is opposite the longest side, \(6\).
    In any triangle the ordering of sides matches the ordering of opposite angles.
  2. Use \(\cos\theta = \dfrac{a^2 + b^2 - c^2}{2ab}\) with \(a = 4\), \(b = 5\), \(c = 6\).
    The two sides adjacent to the angle go on the bottom; the opposite side is the one subtracted on top.
  3. Substitute: \(\dfrac{16 + 25 - 36}{2(4)(5)} = \dfrac{5}{40} = \dfrac18\).
    Positive, so the largest angle is acute - this triangle has no obtuse angle at all.
Where it goes wrong
  • Putting the longest side on the bottom, giving \(\frac{9}{16}\) or similar.
  • Assuming the largest angle must be obtuse and choosing \(-\frac18\).
Takeaway \(\cos\theta = \frac{a^2+b^2-c^2}{2ab}\) with \(c\) opposite \(\theta\); a positive result means an acute angle.
3. TRI-03Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★☆☆

A sector of a circle of radius \(6\) cm has area \(15\pi\) cm\(^2\). Find the arc length of the sector.

  1. A\(\dfrac{5\pi}{6}\) cm
  2. B\(10\pi\) cm
  3. C\(5\pi\) cm
  4. D\(\dfrac{5\pi}{2}\) cm
  5. E\(3\pi\) cm
  6. F\(15\pi\) cm
Answer: C. \(\frac12 r^2\theta = 18\theta = 15\pi \Rightarrow \theta = \frac{5\pi}{6}\). Arc \(= r\theta = 6 \cdot \frac{5\pi}{6} = 5\pi\).
Full method — how and why
Sector area gives the angle; the angle gives the arc length. Both formulas need radians.
  1. Use the sector area formula \(A = \frac12 r^2\theta\): \(\frac12(36)\theta = 18\theta\).
    \(r = 6\), so \(\frac12 r^2 = 18\). The formula is only valid with \(\theta\) in radians.
  2. Set equal to the given area: \(18\theta = 15\pi\), so \(\theta = \dfrac{5\pi}{6}\).
    Divide: \(\frac{15\pi}{18} = \frac{5\pi}{6}\).
  3. Use the arc-length formula \(s = r\theta = 6 \times \dfrac{5\pi}{6} = 5\pi\) cm.
    The \(6\) cancels neatly - a sign the angle is right.
Where it goes wrong
  • Quoting the angle \(\frac{5\pi}{6}\) as the arc length - it is offered as a distractor.
  • Using \(A = r^2\theta\) without the \(\frac12\), which halves the angle and gives \(\frac{5\pi}{2}\).
Takeaway \(A = \frac12 r^2\theta\) and \(s = r\theta\) (radians): find \(\theta\) from one, then substitute into the other.
4. TRI-05Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★☆☆

Convert \(135^\circ\) to radians.

  1. A\(\dfrac{4\pi}{3}\)
  2. B\(\dfrac{3\pi}{8}\)
  3. C\(\dfrac{3\pi}{4}\)
  4. D\(\dfrac{5\pi}{6}\)
  5. E\(\dfrac{2\pi}{3}\)
  6. F\(\dfrac{5\pi}{4}\)
Answer: C. \(135 = \frac34 \times 180\), so \(\frac{3\pi}{4}\).
Full method — how and why
Degree-to-radian conversion: multiply by \(\frac{\pi}{180}\), or spot the fraction of \(180^\circ\).
  1. Express \(135\) as a fraction of \(180\): \(\dfrac{135}{180} = \dfrac34\).
    Since \(180^\circ = \pi\) radians, the same fraction of \(\pi\) gives the answer.
  2. So \(135^\circ = \dfrac34 \pi = \dfrac{3\pi}{4}\).
    Equivalently \(135 \times \frac{\pi}{180} = \frac{3\pi}{4}\).
Where it goes wrong
  • Multiplying by \(\frac{180}{\pi}\) (the reverse conversion).
  • Simplifying \(\frac{135}{180}\) incorrectly to \(\frac23\) or \(\frac56\), both of which are offered.
Takeaway \(180^\circ = \pi\): convert by multiplying by \(\frac{\pi}{180}\), and cancel the fraction fully.
5. TRI-09Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★☆☆

Two sides of a triangle are \(3\) and \(5\) and the included angle is \(\theta\), where \(\cos\theta = -\dfrac{1}{5}\). Find the length of the third side.

  1. A\(\sqrt{34}\)
  2. B\(6\)
  3. C\(\sqrt{40}\)
  4. D\(7\)
  5. E\(\sqrt{28}\)
  6. F\(\sqrt{31}\)
Answer: C. \(c^2 = 9 + 25 - 2\cdot3\cdot5\cdot\left(-\frac15\right) = 34 + 6 = 40\).
Full method — how and why
Cosine rule with a negative cosine. The only difficulty is handling the double negative.
  1. Write the rule: \(c^2 = 3^2 + 5^2 - 2(3)(5)\cos\theta\).
    Two sides and the included angle again.
  2. Substitute \(\cos\theta = -\frac15\): \(c^2 = 9 + 25 - 30\left(-\frac15\right) = 34 + 6\).
    Subtracting a negative adds: \(-30 \times (-\frac15) = +6\).
  3. So \(c^2 = 40\) and \(c = \sqrt{40}\).
    Leave it in surd form; \(\sqrt{40} = 2\sqrt{10}\) if you prefer.
Where it goes wrong
  • Getting \(34 - 6 = 28\) by mishandling the sign, which is offered as \(\sqrt{28}\).
  • Forgetting the angle is obtuse, so the third side must be longer than \(\sqrt{34}\) - a useful sanity check.
Takeaway A negative cosine makes the third side longer than Pythagoras would give; check your answer against \(\sqrt{a^2+b^2}\).
6. TRI-13Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★☆☆

The minute hand of a clock is \(12\) cm long. How far does its tip travel in \(25\) minutes?

  1. A\(25\pi\) cm
  2. B\(5\pi\) cm
  3. C\(10\pi\) cm
  4. D\(12\pi\) cm
  5. E\(\dfrac{25\pi}{6}\) cm
  6. F\(20\pi\) cm
Answer: C. \(25\) minutes is \(\frac{25}{60} = \frac{5}{12}\) of a revolution: \(\frac{5}{12} \times 2\pi \times 12 = 10\pi\) cm.
Full method — how and why
Distance travelled by a rotating tip is arc length. Find the fraction of a full turn, then multiply by the circumference.
  1. Find the fraction of a revolution: \(\dfrac{25}{60} = \dfrac{5}{12}\).
    The minute hand takes \(60\) minutes for one full turn.
  2. Compute the full circumference: \(2\pi r = 2\pi(12) = 24\pi\) cm.
    The tip traces a circle of radius equal to the hand's length.
  3. Multiply: \(\dfrac{5}{12} \times 24\pi = 10\pi\) cm.
    Equivalently, \(\theta = \frac{5}{12}\times 2\pi = \frac{5\pi}{6}\) and \(s = r\theta = 12 \times \frac{5\pi}{6} = 10\pi\).
Where it goes wrong
  • Using the radius as the distance, or the area formula instead of arc length.
  • Using \(\frac{25}{12}\) or \(\frac{25}{60}\) of the radius rather than the circumference.
Takeaway Arc length \(=\) (fraction of a turn) \(\times 2\pi r\); a clock hand makes one turn per hour.
7. X2-15Triangles, radians and sectors · spec MM4.1 · Paper 2 style · ★☆☆

In triangle \(ABC\), \(P\): '\(a^2 = b^2 + c^2\)'. \(Q\): 'angle \(A = 90^\circ\)'. 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: A. Pythagoras and its converse: the cosine rule gives \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\), which is \(0\) exactly when \(a^2 = b^2 + c^2\).
Full method — how and why
Pythagoras and its converse both hold, so this is a genuine equivalence - best seen through the cosine rule.
  1. The cosine rule gives \(a^2 = b^2+c^2-2bc\cos A\), so \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\).
    Rearranging the cosine rule turns both statements into one equation.
  2. \(a^2 = b^2+c^2\) makes the numerator zero, so \(\cos A = 0\) and (in a triangle) \(A = 90^\circ\).
    \(2bc \neq 0\) since side lengths are positive; and \(0^\circ
  3. Conversely \(A = 90^\circ\) gives \(\cos A = 0\), so \(b^2+c^2-a^2 = 0\).
    The same equation, read the other way.
  4. So \(P\) and \(Q\) are equivalent: necessary and sufficient.
    This is Pythagoras together with its converse.
Where it goes wrong
  • Assuming only the forward direction (right angle \(\Rightarrow\) Pythagoras) and answering 'necessary but not sufficient'.
  • Forgetting that \(a\) must be the side opposite \(A\) for the relation to hold.
Takeaway The cosine rule contains Pythagoras and its converse: \(\cos A = 0 \iff a^2 = b^2+c^2\), with \(a\) opposite \(A\).
8. TRI-04Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

Triangle \(PQR\) has area \(10\sqrt{3}\), \(PQ = 8\) and \(PR = 5\). Given that angle \(QPR\) is obtuse, find \(QR\).

  1. A\(13\)
  2. B\(\sqrt{89}\)
  3. C\(\sqrt{89 + 40\sqrt{3}}\)
  4. D\(\sqrt{129}\)
  5. E\(7\)
  6. F\(\sqrt{109}\)
Answer: D. \(\frac12\cdot8\cdot5\sin P = 10\sqrt3 \Rightarrow \sin P = \frac{\sqrt3}{2}\), and \(P\) obtuse gives \(P = 120^\circ\). \(QR^2 = 64 + 25 - 80\cos120^\circ = 89 + 40 = 129\).
Full method — how and why
Area gives \(\sin P\), which has two possible angles. The word 'obtuse' picks the right one, and then the cosine rule finishes.
  1. Use area \(= \frac12 ab\sin C\): \(\frac12(8)(5)\sin P = 10\sqrt3\), so \(20\sin P = 10\sqrt3\) and \(\sin P = \dfrac{\sqrt3}{2}\).
    The formula uses the angle between the two sides, which is \(QPR\) here.
  2. \(\sin P = \frac{\sqrt3}{2}\) gives \(P = 60^\circ\) or \(P = 120^\circ\); the obtuse condition selects \(P = 120^\circ\).
    \(\sin\theta = \sin(180^\circ - \theta)\): a sine value never determines the angle by itself.
  3. Apply the cosine rule with \(\cos 120^\circ = -\frac12\): \(QR^2 = 64 + 25 - 2(8)(5)\left(-\frac12\right) = 89 + 40 = 129\).
    A negative cosine turns the subtraction into an addition, making the opposite side longer - which is what an obtuse angle should do.
  4. So \(QR = \sqrt{129}\).
    Left in surd form; no calculator is available anyway.
Where it goes wrong
  • Taking \(P = 60^\circ\) and answering \(7\) - the value the question's 'obtuse' is designed to exclude.
  • Using \(\cos 120^\circ = +\frac12\), giving \(\sqrt{49}\).
Takeaway \(\sin\theta\) always allows two angles; use any stated acute/obtuse condition, and remember \(\cos\) is negative for obtuse angles.
9. TRI-06Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★★☆

A chord of a circle of radius \(4\) subtends an angle of \(\dfrac{\pi}{3}\) at the centre. Find the area of the minor segment cut off by the chord.

  1. A\(\dfrac{8\pi}{3} - 8\sqrt{3}\)
  2. B\(\dfrac{8\pi}{3} - 4\)
  3. C\(\dfrac{16\pi}{3} - 4\sqrt3\)
  4. D\(8\pi - 4\sqrt3\)
  5. E\(\dfrac{8\pi}{3} - 4\sqrt{3}\)
  6. F\(\dfrac{4\pi}{3} - 2\sqrt3\)
Answer: E. Sector area \(\frac12 \cdot 16 \cdot \frac\pi3 = \frac{8\pi}{3}\). Triangle area \(\frac12 \cdot 16 \sin\frac\pi3 = 4\sqrt3\). Segment \(= \frac{8\pi}{3} - 4\sqrt3\).
Full method — how and why
Segment \(=\) sector \(-\) triangle. Compute the two areas and subtract; both use the same \(r\) and \(\theta\).
  1. Sector area: \(\frac12 r^2\theta = \frac12(16)\left(\frac{\pi}{3}\right) = \dfrac{8\pi}{3}\).
    \(r = 4\) so \(r^2 = 16\); the angle is already in radians.
  2. Triangle area (two radii and the chord): \(\frac12 r^2\sin\theta = \frac12(16)\sin\frac{\pi}{3} = 8 \times \dfrac{\sqrt3}{2} = 4\sqrt3\).
    Use \(\frac12 ab\sin C\) with both sides equal to the radius.
  3. Subtract: segment \(= \dfrac{8\pi}{3} - 4\sqrt3\).
    The minor segment is the bit of the sector left over once the triangle is removed.
Where it goes wrong
  • Forgetting the \(\frac12\) in the triangle area, giving \(\frac{8\pi}{3} - 8\sqrt3\) (an offered option), which is negative.
  • Using \(\sin\frac{\pi}{3} = \frac12\).
Takeaway Segment area \(= \frac12 r^2(\theta - \sin\theta)\) - sector minus triangle, all in radians.
10. TRI-07Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

In triangle \(ABC\), \(AB = 6\), \(BC = 4\) and angle \(BAC = 30^\circ\). How many possible triangles are there?

  1. A\(1\)
  2. B\(3\)
  3. C\(2\)
  4. D\(0\)
  5. Einfinitely many
Answer: C. Sine rule: \(\dfrac{\sin C}{6} = \dfrac{\sin 30^\circ}{4}\), so \(\sin C = \frac34\). Since \(BC = 4 < AB = 6\) and \(4 > 6\sin30^\circ = 3\), both the acute and obtuse values of \(C\) give valid triangles: the ambiguous case, two triangles.
Full method — how and why
Side-side-angle data: the ambiguous case. Compute \(\sin C\), then decide whether both the acute and obtuse angles give a valid triangle.
  1. Apply the sine rule: \(\dfrac{\sin C}{AB} = \dfrac{\sin A}{BC}\), i.e. \(\dfrac{\sin C}{6} = \dfrac{\sin 30^\circ}{4}\).
    Pair each angle with the side opposite it: \(C\) is opposite \(AB = 6\), and \(A\) is opposite \(BC = 4\).
  2. Solve: \(\sin C = \dfrac{6 \times 0.5}{4} = \dfrac34\), which is less than \(1\), so solutions exist.
    If \(\sin C\) came out greater than \(1\), no triangle would exist at all.
  3. Both \(C = \arcsin\frac34\) and \(180^\circ - \arcsin\frac34\) are possible in principle; check whether the obtuse one still leaves a positive third angle.
    \(\arcsin\frac34 \approx 48.6^\circ\), so the obtuse option is about \(131.4^\circ\); with \(A = 30^\circ\) that totals under \(180^\circ\) ✓.
  4. Equivalently: the side opposite the given angle (\(4\)) is shorter than the other given side (\(6\)) but longer than \(6\sin30^\circ = 3\) - the classic two-triangle condition. So there are 2 triangles.
    The condition \(b\sin A < a < b\) is the standard test for the ambiguous case.
Where it goes wrong
  • Assuming one triangle because the sine rule 'gives one answer'.
  • Assuming two triangles always - if \(a \geq b\) the obtuse option would make the angles exceed \(180^\circ\).
Takeaway SSA data: with \(b\sin A < a < b\) there are two triangles; with \(a \geq b\) exactly one; with \(a < b\sin A\) none.
11. TRI-08Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★★☆

The perimeter of a sector of radius \(r\) and angle \(\theta\) radians is \(20\), and its area is \(16\). Find the possible values of \(r\).

  1. A\(4\) only
  2. B\(8\) only
  3. C\(1\) or \(16\)
  4. D\(2\) or \(8\)
  5. E\(4\) or \(8\)
  6. F\(2\) or \(4\)
Answer: D. \(2r + r\theta = 20\) and \(\frac12 r^2\theta = 16\). From the first \(r\theta = 20 - 2r\); then \(\frac12 r(20 - 2r) = 16 \Rightarrow r^2 - 10r + 16 = 0 \Rightarrow r = 2\) or \(8\).
Full method — how and why
Two sector facts, two unknowns. Eliminate \(\theta\) by substitution and solve the resulting quadratic in \(r\).
  1. Perimeter of a sector \(=\) two radii \(+\) arc: \(2r + r\theta = 20\).
    Students often forget the two straight edges - the perimeter is not just the arc.
  2. Rearrange for the arc: \(r\theta = 20 - 2r\).
    Keeping the product \(r\theta\) together is the key, because the area formula contains it.
  3. Area: \(\frac12 r^2\theta = \frac12 r(r\theta) = \frac12 r(20 - 2r) = 16\).
    Writing \(r^2\theta\) as \(r \times (r\theta)\) lets you substitute directly - no need to find \(\theta\).
  4. Simplify and solve: \(10r - r^2 = 16 \Rightarrow r^2 - 10r + 16 = 0 \Rightarrow (r-2)(r-8) = 0\), so \(r = 2\) or \(8\).
    Both are geometrically valid: \(r = 2\) gives \(\theta = 8\) radians... which exceeds \(2\pi\), but the question asks only for the values of \(r\) from the equations.
Where it goes wrong
  • Using perimeter \(= r\theta\) alone, which gives a different quadratic.
  • Finding only one root of the quadratic.
Takeaway Substitute the product \(r\theta\) as a block; sector perimeter is \(2r + r\theta\).
12. TRI-10Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

In triangle \(ABC\), \(\sin A : \sin B : \sin C = 3 : 5 : 7\). Find angle \(C\).

  1. A\(120^\circ\)
  2. B\(150^\circ\)
  3. C\(100^\circ\)
  4. D\(135^\circ\)
  5. E\(90^\circ\)
  6. F\(60^\circ\)
Answer: A. By the sine rule the sides are in ratio \(3 : 5 : 7\). \(\cos C = \dfrac{9 + 25 - 49}{2\cdot3\cdot5} = -\dfrac{15}{30} = -\dfrac12\), so \(C = 120^\circ\).
Full method — how and why
A ratio of sines is a ratio of sides (sine rule). Then the cosine rule finds the angle.
  1. By the sine rule, \(a : b : c = \sin A : \sin B : \sin C = 3 : 5 : 7\).
    \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) means sides are proportional to the sines of opposite angles.
  2. Take the sides as \(3k, 5k, 7k\); the \(k\) will cancel, so use \(3, 5, 7\).
    Only the shape matters for angles, so the scale factor is irrelevant.
  3. Apply the cosine rule for the angle opposite \(7\): \(\cos C = \dfrac{9 + 25 - 49}{2(3)(5)} = \dfrac{-15}{30} = -\dfrac12\).
    \(C\) is opposite the largest side, so expect the largest angle.
  4. \(\cos C = -\frac12\) gives \(C = 120^\circ\).
    One of the exact values: \(\cos 120^\circ = -\cos 60^\circ = -\frac12\).
Where it goes wrong
  • Treating \(3:5:7\) as the angles themselves, giving \(84^\circ\).
  • Reading \(\cos C = -\frac12\) as \(60^\circ\) and ignoring the sign.
Takeaway Sines in ratio \(\Rightarrow\) sides in the same ratio; then use the cosine rule with any convenient multiple.
13. TRI-11Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★★☆

A circle has radius \(r\). A sector of angle \(\theta\) radians has the same area as the segment cut off by a chord that subtends an angle of \(\dfrac{\pi}{2}\) at the centre. Find \(\theta\).

  1. A\(\dfrac{\pi}{2} - 1\)
  2. B\(\dfrac{\pi}{4}\)
  3. C\(\dfrac{\pi}{4} - \dfrac12\)
  4. D\(1\)
  5. E\(\pi - 2\)
  6. F\(\dfrac{\pi}{2} - \dfrac12\)
Answer: A. Segment area \(= \frac12 r^2\left(\frac\pi2 - \sin\frac\pi2\right) = \frac12 r^2\left(\frac\pi2 - 1\right)\). Sector area \(\frac12 r^2\theta\). So \(\theta = \frac\pi2 - 1\).
Full method — how and why
Equate two area expressions. The radius cancels, leaving an equation for \(\theta\) alone.
  1. Segment area for a right angle at the centre: \(\frac12 r^2\left(\frac{\pi}{2} - \sin\frac{\pi}{2}\right) = \frac12 r^2\left(\frac{\pi}{2} - 1\right)\).
    Segment \(=\) sector \(-\) triangle \(= \frac12 r^2\theta - \frac12 r^2\sin\theta\), and \(\sin\frac\pi2 = 1\).
  2. Sector area for angle \(\theta\): \(\frac12 r^2\theta\).
    Same radius, different angle.
  3. Set them equal: \(\frac12 r^2\theta = \frac12 r^2\left(\frac\pi2 - 1\right)\).
    Both areas are in the same circle, so \(r\) is common.
  4. Cancel \(\frac12 r^2\): \(\theta = \dfrac{\pi}{2} - 1\).
    The answer is independent of \(r\), which is why the question never tells you its value.
Where it goes wrong
  • Using \(\sin\frac\pi2 = 0\) and answering \(\frac\pi2\).
  • Forgetting to cancel and leaving an \(r\) in the answer.
Takeaway When two areas in the same circle are equated, \(r\) cancels - the answer depends only on the angles.
14. TRI-12Triangles, radians and sectors · spec MM4.1 · Paper 2 style · ★★☆

A triangle has sides \(a\), \(b\), \(c\) with \(a^2 + b^2 < c^2\). Which of the following statements is/are true?

  1. IThe angle opposite \(c\) is obtuse.
  2. II\(c\) is the longest side.
  3. IIIThe triangle cannot be isosceles.
  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: E. Cosine rule: \(\cos C = \frac{a^2 + b^2 - c^2}{2ab} < 0\), so \(C\) is obtuse (I). The largest angle is opposite the longest side, so \(c\) is longest (II). III is false: \(a = b = 1\), \(c = 1.5\) works.
Full method — how and why
A condition on the sides. The cosine rule converts it into a statement about an angle; then test the third claim with a counterexample.
  1. I: \(\cos C = \dfrac{a^2+b^2-c^2}{2ab}\), and \(a^2 + b^2 < c^2\) makes the numerator negative, so \(\cos C < 0\) and \(C\) is obtuse. True.
    The denominator \(2ab\) is positive, so the sign of \(\cos C\) is the sign of \(a^2 + b^2 - c^2\).
  2. II: a triangle can have at most one obtuse angle, and the largest angle is opposite the largest side. Since \(C\) is obtuse it is the largest angle, so \(c\) is the longest side. True.
    Two obtuse angles would already exceed \(180^\circ\).
  3. III: try \(a = b = 1\) and \(c = 1.5\): then \(a^2 + b^2 = 2 < 2.25 = c^2\) ✓ and the triangle is isosceles. False.
    A single counterexample kills the claim. Check the triangle inequality too: \(1 + 1 > 1.5\) ✓.
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Assuming an obtuse triangle cannot be isosceles.
  • Forgetting to check the triangle inequality when constructing a counterexample.
Takeaway \(a^2+b^2 < c^2\) \(\Leftrightarrow\) the angle opposite \(c\) is obtuse; obtuse triangles can still be isosceles.
15. TRI-14Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

A vertical pole \(BT\) stands on horizontal ground. From a point \(A\) on the ground the angle of elevation of \(T\) is \(30^\circ\); from a point \(C\) on the ground, \(20\) m closer to the pole than \(A\) and in line with \(A\) and \(B\), the angle of elevation is \(60^\circ\). Find the height of the pole.

  1. A\(10\) m
  2. B\(20\sqrt{3}\) m
  3. C\(15\) m
  4. D\(10\sqrt{3}\) m
  5. E\(\dfrac{20}{\sqrt3}\) m
  6. F\(30\) m
Answer: D. Let the height be \(h\). \(AB = h\sqrt3\) and \(CB = \frac{h}{\sqrt3}\). \(AB - CB = 20 \Rightarrow h\left(\sqrt3 - \frac1{\sqrt3}\right) = \frac{2h}{\sqrt3} = 20 \Rightarrow h = 10\sqrt3\).
Full method — how and why
Two right-angled triangles sharing the same height. Express both base lengths in terms of \(h\) and use the given difference.
  1. From \(A\): \(\tan 30^\circ = \dfrac{h}{AB}\), so \(AB = \dfrac{h}{\tan 30^\circ} = h\sqrt3\).
    \(\tan 30^\circ = \frac{1}{\sqrt3}\), so dividing by it multiplies by \(\sqrt3\).
  2. From \(C\): \(\tan 60^\circ = \dfrac{h}{CB}\), so \(CB = \dfrac{h}{\sqrt3}\).
    \(\tan 60^\circ = \sqrt3\). The steeper angle means the shorter base - a good sanity check.
  3. Use the \(20\) m gap: \(AB - CB = h\sqrt3 - \dfrac{h}{\sqrt3} = h\left(\dfrac{3 - 1}{\sqrt3}\right) = \dfrac{2h}{\sqrt3} = 20\).
    Common denominator \(\sqrt3\): \(\sqrt3 = \frac{3}{\sqrt3}\).
  4. Solve: \(h = \dfrac{20\sqrt3}{2} = 10\sqrt3\) m.
    Multiply both sides by \(\sqrt3\) and halve.
Where it goes wrong
  • Adding the bases instead of subtracting (the closer point has the shorter base).
  • Using \(\tan 30^\circ = \sqrt3\), which swaps the two triangles.
Takeaway Two elevation angles to the same object: write each horizontal distance in terms of the height and use their difference.
16. TRI-15Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★★☆

Triangle \(ABC\) has \(AB = 7\), \(BC = 8\), \(CA = 9\). Find the area of the triangle.

  1. A\(28\)
  2. B\(\sqrt{720}\)
  3. C\(24\sqrt5\)
  4. D\(12\sqrt{5}\)
  5. E\(30\)
  6. F\(6\sqrt{5}\)
Answer: D. \(\cos A = \dfrac{49 + 81 - 64}{2\cdot7\cdot9} = \dfrac{66}{126} = \dfrac{11}{21}\), so \(\sin A = \dfrac{\sqrt{441 - 121}}{21} = \dfrac{\sqrt{320}}{21} = \dfrac{8\sqrt5}{21}\). Area \(= \frac12\cdot7\cdot9\cdot\frac{8\sqrt5}{21} = 12\sqrt5\).
Full method — how and why
Three sides, area wanted. Find a cosine, convert to a sine, then use \(\frac12 ab\sin C\) - all without a calculator.
  1. Find \(\cos A\) where \(A\) is between the sides \(7\) and \(9\): \(\cos A = \dfrac{49 + 81 - 64}{2(7)(9)} = \dfrac{66}{126} = \dfrac{11}{21}\).
    The side opposite \(A\) is \(BC = 8\), so that is the one subtracted.
  2. Convert to \(\sin A\) using \(\sin^2 + \cos^2 = 1\): \(\sin A = \sqrt{1 - \frac{121}{441}} = \dfrac{\sqrt{320}}{21}\).
    Keep the denominator \(21\) throughout: \(1 = \frac{441}{441}\).
  3. Simplify the surd: \(\sqrt{320} = \sqrt{64 \times 5} = 8\sqrt5\), so \(\sin A = \dfrac{8\sqrt5}{21}\).
    Pull out the largest square factor.
  4. Area \(= \frac12(7)(9)\sin A = \frac{63}{2} \times \dfrac{8\sqrt5}{21} = 12\sqrt5\).
    \(\frac{63 \times 8}{2 \times 21} = \frac{504}{42} = 12\).
Where it goes wrong
  • Taking \(\sin A\) positive is right (angles in a triangle are between \(0^\circ\) and \(180^\circ\)), but forgetting to square \(\cos A\) before subtracting.
  • Arithmetic slips in \(\sqrt{320}\), e.g. \(16\sqrt5\), doubling the area.
Takeaway Three sides \(\to\) \(\cos\) from the cosine rule \(\to\) \(\sin\) from the Pythagorean identity \(\to\) area from \(\frac12 ab\sin C\).
17. TRI-16Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★★★

Two circles of radius \(2\) have their centres a distance \(2\) apart. Find the area of the region common to both circles.

  1. A\(\dfrac{2\pi}{3}\)
  2. B\(\dfrac{4\pi}{3} - 2\sqrt3\)
  3. C\(\dfrac{8\pi}{3} - 2\sqrt{3}\)
  4. D\(\dfrac{8\pi}{3} - 4\sqrt3\)
  5. E\(2\pi - 2\sqrt3\)
  6. F\(\dfrac{4\pi}{3} - \sqrt3\)
Answer: C. The common chord subtends \(\frac{2\pi}{3}\) at each centre (equilateral triangles of side \(2\)). The overlap is two segments: \(2 \times \frac12\cdot4\left(\frac{2\pi}{3} - \sin\frac{2\pi}{3}\right) = 4\left(\frac{2\pi}{3} - \frac{\sqrt3}{2}\right) = \frac{8\pi}{3} - 2\sqrt3\).
Full method — how and why
The overlap of two circles is two segments. The geometry of the centres gives the angle, then the segment formula applies twice.
  1. Let the circles meet at \(P\) and \(Q\). Each centre, together with \(P\) and \(Q\), forms a triangle with sides \(2\), \(2\) and the chord \(PQ\).
    Both radii are \(2\) and the centres are \(2\) apart, so joining a centre to \(P\) and to the other centre gives an equilateral triangle of side \(2\).
  2. That equilateral triangle shows the half-angle at a centre is \(60^\circ\), so the chord subtends \(\frac{2\pi}{3}\) (i.e. \(120^\circ\)) at each centre.
    Two equilateral triangles back to back, one for \(P\) and one for \(Q\).
  3. Compute one segment: \(\frac12 r^2(\theta - \sin\theta) = \frac12(4)\left(\frac{2\pi}{3} - \frac{\sqrt3}{2}\right) = \frac{4\pi}{3} - \sqrt3\).
    \(\sin\frac{2\pi}{3} = \frac{\sqrt3}{2}\).
  4. The overlap consists of two such segments (one from each circle): \(2\left(\frac{4\pi}{3} - \sqrt3\right) = \dfrac{8\pi}{3} - 2\sqrt3\).
    By symmetry the two segments are congruent.
Where it goes wrong
  • Counting only one segment, giving \(\frac{4\pi}{3} - \sqrt3\) - an offered option.
  • Using \(\theta = \frac{\pi}{3}\) (the half-angle) instead of \(\frac{2\pi}{3}\).
Takeaway Overlapping circles: the lens is two segments; get the central angle from the triangle formed by the two centres and an intersection point.