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IB DP Maths AA · Topic 3 Geometry and trigonometry · SL 3.1–3.4 · ~7 min read

Trigonometry: triangles, radians, arcs and sectors

Distance and midpoint in 3D, volume and surface area of solids, angles in 3D, right-angled trig, the sine rule (including the ambiguous case), the cosine rule, area ½ab sin C, bearings, radians, arc length and sector area.

3D distance & midpointVolume & surface areaAngles in 3DSine ruleAmbiguous caseCosine ruleArea ½ab sin CBearings
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Right-angled triangles: SOH CAH TOA. Any triangle: sine rule for "pairs" (a side and its opposite angle known), cosine rule for SAS or SSS.
  • Area of any triangle = ½ab sin C (two sides and the angle between them).
  • The ambiguous case: given two sides and a non-included angle, sin B = k can give two angles, B and 180° − B. Check both.
  • Radians: π rad = 180°. With θ in radians, arc length = rθ and sector area = ½r2θ.
  • This note is the same at SL and HL, and unchanged for 2029.

By the end you can

  • Find distances, midpoints, volumes and surface areas of 3D solids.
  • Find angles between lines and between a line and a plane in 3D.
  • Use the sine rule, including the ambiguous case, and the cosine rule.
  • Find the area of a triangle with ½ab sin C, and solve bearings problems.
  • Convert between degrees and radians and find arc lengths, sector areas and segment areas.

13D geometry

d = √((x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2) FORMULA BOOKLET
distance between two points in 3D
midpoint = (x1 + x22, y1 + y22, z1 + z22) FORMULA BOOKLET
average each coordinate
SolidVolumeSurface area
Sphere43πr34πr2
Cone13πr2hπrl (curved), l = slant height
Pyramid13 × base area × hadd the faces
Prism / cylinderbase area × h2πrh + 2πr2 (cylinder)
  • All of these are in the formula booklet. Hemispheres and composite solids come up often: add or subtract the pieces, and remember the flat face of a hemisphere.
θABCG6412angle between AGand the base ABCD:tan θ = CG/ACAC: base diagonal (blue)CG: height (green)
To find the angle between a line and a plane, drop a perpendicular to the plane and use the right-angled triangle that forms.
Worked example

A cuboid has base 6 cm by 4 cm and height 12 cm. Find the angle between the diagonal AG and the base.

  1. base diagonal AC = √(62 + 42) = √52 M1
  2. tan θ = 12/√52 M1
  3. θ = 59.0° A1

2Sine rule, cosine rule and area

ABCcaba/sin A = b/sin B = c/sin Cc² = a² + b² − 2ab cos Carea = ½ab sin CSide a is opposite angle A, and so on. All three results are in the formula booklet.
Label sides with lower-case letters opposite the matching capital-letter angles.
asin A = bsin B = csin C FORMULA BOOKLET
sine rule: use with a side–angle pair
c2 = a2 + b2 − 2ab cos C, cos C = a2 + b2 − c22ab FORMULA BOOKLET
cosine rule: SAS or SSS
A = ½ab sin C FORMULA BOOKLET
area of a triangle
Worked example

In triangle ABC, A = 40°, B = 65° and a = 8 cm. Find b.

  1. bsin 65° = 8sin 40° M1
  2. b = 11.3 cm A1
Worked example

In triangle ABC, a = 7 cm, b = 9 cm and C = 50°. Find c and the area.

  1. c2 = 49 + 81 − 126 cos 50° M1
  2. c = 7.00 cm A1
  3. area = ½(7)(9) sin 50° = 24.1 cm2 A1
Worked example

A triangle has sides 5, 7 and 9 cm. Find its largest angle.

  1. the largest angle is opposite the longest side R1
  2. cos θ = 25 + 49 − 812 × 5 × 7 = −0.1 M1
  3. θ = 95.7° A1
Common mistakeUsing the sine rule to find the largest angle of a triangle. The sine rule cannot tell 84° from 96°; the cosine rule can, because cos is negative for obtuse angles.

3The ambiguous case

ACB₁B₂40°Given A, b and a (SSA): side a can swing to two positionstwo triangles: B = 55.5° or 124.5°
With angle A and sides b and a fixed, side a can meet the base in two places.
Worked example

In triangle ABC, A = 40°, a = 11.7 cm and b = 15 cm. Find the possible values of B.

  1. sin B = 15 sin 40°11.7 = 0.824 M1
  2. B = 55.5° A1
  3. or B = 180° − 55.5° = 124.5°; both work because 124.5° + 40° < 180° A1R1
Exam tipTwo triangles are possible only when the given angle is opposite the shorter of the two given sides. If it is opposite the longer side, there is just one triangle.

4Bearings

  • Bearings are measured clockwise from North and written with three figures: 060°, 150°, 305°.
Worked example

A ship sails 12 km on a bearing of 060°, then 8 km on a bearing of 150°. Find its distance and bearing from the start.

  1. the turn is 150° − 60° = 90°, so the triangle is right-angled R1
  2. distance = √(144 + 64) = 14.4 km A1
  3. angle at the start = tan−1(8/12) = 33.7°, so bearing = 60° + 33.7° = 094° M1A1
Exam tipAlways draw a North line at every point. Co-interior angles between parallel North lines add to 180°, which gives the angle inside the triangle.

5Radians, arcs and sectors

θrarc lθ in radiansarc length l = rθsector area = ½r²θsegment (red) =½r²(θ − sin θ)π rad = 180°
In radians the formulas lose their "÷ 360" and become short.
l = rθ, A = ½r2θ FORMULA BOOKLET
θ in radians
  • Converting: multiply degrees by π/180; multiply radians by 180/π. Learn 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2.
Worked example

A sector has radius 5 cm and angle 0.8 rad. Find its perimeter and area.

  1. arc = 5 × 0.8 = 4 cm A1
  2. perimeter = 4 + 5 + 5 = 14 cm A1
  3. area = ½(25)(0.8) = 10 cm2 A1
Worked example

Find the area of the segment cut off by a chord that subtends π/3 at the centre of a circle of radius 6 cm.

  1. sector − triangle = ½(36)(π/3) − ½(36) sin(π/3) M1
  2. = 6π − 9√3 A1
  3. = 3.26 cm2 A1
Common mistakeLeaving the calculator in degree mode when a question is in radians (or the reverse). In Paper 1 and 2, check the mode at the start of every trig question.
GDC (calculator) methodPaper 2: the GDC solves triangles fast, but write the rule you used and the substituted values first, or you lose the method marks.
Linking questionThe unit circle extends sin, cos and tan to all angles (Trig identities, equations and graphs). Radians are needed for calculus, because d/dx (sin x) = cos x only in radians (Differentiation).
Idea for your IA / explorationUse trigonometry to measure something you can't reach (the height of a building from two angles of elevation) and compare with the real value, or model the geometry of a sports pitch.

Test yourself — cover the answers first

Quick check

  1. Convert 150° to radians.
  2. Find the arc length for r = 10 cm, θ = 0.3 rad.
  3. Which rule finds an angle from three sides?
  4. Find the area of a triangle with sides 4 and 6 cm and an included angle of 30°.
  5. When can the ambiguous case happen?

Exam-style

  1. A sector of radius r and angle θ has perimeter 20 cm and area 16 cm². Find r and θ.
  2. Points A(1, 2, −1) and B(5, −2, 1). Find AB and the midpoint of AB.
Show answers
  1. 5π/6
  2. 3 cm
  3. the cosine rule
  4. 6 cm²
  5. two sides and a non-included angle are given
  6. 2r + rθ = 20 and ½r²θ = 16
    rθ = 20 − 2r, so ½r(20 − 2r) = 16 ⇒ r² − 10r + 16 = 0
    r = 2, θ = 8 or r = 8, θ = 0.5 (θ = 8 rad is more than 2π, so reject it if θ must be less than 2π)
  7. AB = √(16 + 16 + 4) = 6
    midpoint (3, 0, 0)
Practise this chapter
IB Maths AA practice papers
Next chapterTrig identities, equations and graphs →
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.