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Triangles, radians and sectors

Revision notes · spec MM4.1, MM4.2, M5.16

Section 1 Part 1: AS-level pure mathematics

Sine rule, cosine rule and \(\frac12 ab\sin C\) solve any triangle; radians make arc length and sector area simple (\(r\theta\) and \(\frac12 r^2\theta\)). The 'ambiguous case' - two triangles from side-side-angle data - is a TMUA favourite.

Triangles (MM4.1)

  • Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\). Cosine rule: \(a^2 = b^2 + c^2 - 2bc\cos A\), i.e. \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\).
  • Area \(= \frac12 ab\sin C\). Heron-style: find \(\cos\) from the cosine rule, then \(\sin\) from \(\sin^2 + \cos^2 = 1\).
  • Ambiguous case: given sides \(a, b\) and angle \(A\) (opposite \(a\)) with \(b\sin A < a < b\), there are two triangles (an acute and an obtuse value of \(B\)). If \(a \geq b\) or \(a = b\sin A\) there is one; if \(a < b\sin A\), none.
  • Largest angle is opposite the longest side; \(a^2 + b^2 < c^2\) means the angle opposite \(c\) is obtuse.

Radians, arcs and sectors (MM4.2)

  • \(\pi\) radians \(= 180^\circ\); \(30^\circ = \frac\pi6\), \(45^\circ = \frac\pi4\), \(60^\circ = \frac\pi3\), \(90^\circ = \frac\pi2\).
  • Arc length \(s = r\theta\); sector area \(\frac12 r^2\theta\); segment area \(\frac12 r^2(\theta - \sin\theta)\) (\(\theta\) in radians).
  • Perimeter of a sector \(= 2r + r\theta\).
  • Overlapping circles: the common region is two segments; the chord subtends \(2\theta\) at each centre.

Must-know (no formula booklet)

  • \(a^2 = b^2 + c^2 - 2bc\cos A\)
  • \(\frac{a}{\sin A} = \frac{b}{\sin B}\)
  • area \(= \frac12 ab\sin C\)
  • \(s = r\theta\), \(A = \frac12 r^2\theta\)
  • segment \(= \frac12 r^2(\theta - \sin\theta)\)

Classic traps

  • \(\cos 120^\circ = -\frac12\): the cosine rule with an obtuse angle adds the last term.
  • Sector formulas need radians; convert first.
  • An area of \(10\sqrt3\) with \(\sin P = \frac{\sqrt3}{2}\) gives \(P = 60^\circ\) or \(120^\circ\) - read whether the angle is obtuse.

Worked example

Q. Sector: perimeter \(20\), area \(16\). Find \(r\).

A. \(r\theta = 20 - 2r\) and \(\frac12 r(20 - 2r) = 16\), so \(r^2 - 10r + 16 = 0\): \(r = 2\) or \(8\).

How it has been examined

2023 · Paper 1Question 8'fun' triangles: sine rule ambiguous casequestion papersit online
2022 · Paper 1Question 4similar sectors: arc, area and perimeterquestion papersit online
2022 · Paper 1Question 14triangle in a circle of radius 6question papersit online
2022 · Paper 1Question 17two non-congruent triangles: the ambiguous case with sides x - 1 and sqrt(-x^2 + 6x - 5)question papersit online
2021 · Paper 2Question 18can the triangle be drawn?question papersit online
October 2018 · Paper 1Question 19two possible triangles: the larger onequestion papersit online
October 2018 · Paper 2Question 14values of p that determine PQ uniquelyquestion papersit online
November 2017 · Paper 1Question 20range of an angle in a triangle with sides a, a + d, a + 2dquestion papersit online