Every triangle and circle-sector method for IB Mathematics Applications & Interpretation HL — SOHCAHTOA, the sine & cosine rules, area ½ab sinC, arc length & sector area, bearings and 3-D angles. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Trigonometry is the measurement engine of IB Mathematics Applications & Interpretation HL. AI treats it less as abstract identity-juggling and more as a modelling tool: you turn a real situation — a land survey, a ship's journey, a roof truss, a radar sweep — into a triangle or a circle-sector and then measure it. Because a calculator is always allowed and the whole topic runs in degrees, the marks live in choosing the right rule (right-angle ratio vs sine rule vs cosine rule) and reading the diagram correctly, not in grinding out the arithmetic.
This cheatsheet condenses the whole of SL 3.1–3.4 — SOHCAHTOA, the sine and cosine rules with the ambiguous case, the area rule $\tfrac12 ab\sin C$, arc length and sector area, bearings, and 3-D line-and-plane angles — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Right-angled triangles: SOHCAHTOA SL 3.2
In a right-angled triangle the three ratios link an angle to two sides. Label the sides relative to the angle you are using first — opposite, adjacent, hypotenuse — then pick the ratio.
Name the sides relative to θ: the hypotenuse always faces the right angle (the longest side), the opposite faces θ, and the adjacent is the remaining side next to θ.
Sine (SOH):$\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$ — so a height along a slope is $\text{hyp}\times\sin\theta$
Cosine (CAH):$\cos\theta=\dfrac{\text{adj}}{\text{hyp}}$ — the horizontal distance is $\text{hyp}\times\cos\theta$
Find the angle:use the inverse, e.g. $\theta=\tan^{-1}\!\left(\dfrac{\text{opp}}{\text{adj}}\right)$; which two sides you have decides $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$
Pythagoras:$a^2+b^2=c^2$, where $c$ is the hypotenuse
TrickCircle the angle first, then ask which side is opposite it. A length along the slope (ladder, cable, string) is the hypotenuse; the height is opposite the ground angle, so use sin; the base distance uses cos.
TrapThe hypotenuse is fixed — it always faces the right angle — but opposite and adjacent swap the moment you switch to the other acute angle. Re-label before every calculation, and keep your GDC in degree mode.
§2 — The sine rule SL 3.2
Once a triangle is not right-angled, the sine rule takes over — use it whenever you have a side together with the angle opposite it, plus one more piece.
Sine rule:$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ — each side over the sine of its opposite angle
Find a side:$b=a\cdot\dfrac{\sin B}{\sin A}$
Find an angle:flip it: $\dfrac{\sin A}{a}=\dfrac{\sin B}{b}$, so $\sin A=\dfrac{a\sin B}{b}$
TrickAlways pair a side with the angle facing it, never an adjacent one. If your answer looks far too big or too small, you have probably flipped the fraction — swap numerator and denominator.
TrapThe sine rule needs a matched side–angle pair. If all you have is two sides and the angle between them, the sine rule cannot start — that is cosine-rule territory (§4).
§3 — The ambiguous case (SSA) SL 3.3
When you use the sine rule to find an angle from two sides and a non-included angle, the data can describe two different triangles — because a sine value has two angles in $0^\circ$–$180^\circ$.
Two angles, one sine:$\sin\theta=\sin(180^\circ-\theta)$, so an acute solution $\theta$ has an obtuse partner $180^\circ-\theta$
Test it:keep the obtuse option only if the three angles still sum to less than $180^\circ$
NoteAfter the calculator gives the acute angle, always compute $180^\circ$ minus it and check whether that second triangle is still possible. Many exam questions are worth a mark for spotting both answers.
TrapThe ambiguity only appears when you use the sine rule to find an angle. Finding a side with the sine rule, or using the cosine rule for an angle, is never ambiguous — the cosine rule returns a single, correctly-signed answer.
§4 — The cosine rule SL 3.2
The cosine rule handles the two cases the sine rule cannot: two sides and the included angle (find the third side), and three sides (find any angle). It is Pythagoras with a correction term.
Find a side:$a^2=b^2+c^2-2bc\cos A$, where $A$ is the angle between $b$ and $c$
Find an angle:$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ — the lone side $a$ in the numerator faces the angle $A$
NoteWhen $A=90^\circ$, $\cos A=0$ and the $-2bc\cos A$ term vanishes, leaving $a^2=b^2+c^2$: the cosine rule generalises Pythagoras, and the extra term is the correction for the angle not being a right angle.
TrapThe angle must be the one squeezed between the two known sides. Keep the minus sign attached to $2bc\cos A$, and take the square root only at the very end — squaring, then subtracting, then rooting, in that order.
§5 — Area of a triangle: $\tfrac12 ab\sin C$ SL 3.2
With two sides and the angle between them you can get the area straight away — no perpendicular height needed. This is the AI workhorse for triangular plots of land, sails and panels.
Area rule:$\text{Area}=\tfrac12 ab\sin C$, where $C$ is the angle enclosed by sides $a$ and $b$
TrickOnly have three sides? Find one angle with the cosine rule first, then feed it into $\tfrac12 ab\sin C$. Two sides and their included angle is all this formula ever needs.
TrapThe angle must be the included angle between the two sides you multiply, or the formula does not apply. And do not drop the $\tfrac12$: if you compute only $ab\sin C$ you are out by a factor of two.
§6 — Arc length & sector area SL 3.4
A sector is simply a fraction of a whole circle — the fraction $\dfrac{\theta}{360}$ of it, where $\theta$ is the central angle in degrees.
Arc length:$\ell=\dfrac{\theta}{360}\times 2\pi r$ — the fraction $\dfrac{\theta}{360}$ of the circumference $2\pi r$
Sector area:$A=\dfrac{\theta}{360}\times \pi r^2$ — the same fraction of the full area $\pi r^2$
NoteIf the angle is given in radians instead, the fraction disappears: arc length $=r\theta$ and sector area $=\tfrac12 r^2\theta$, with no degree conversion. Use whichever matches the units in the question.
TrapKeep the two formulas apart: arc length pairs with the circumference $2\pi r$ (units of length), sector area pairs with $\pi r^2$ (units of length squared). Mixing them is the classic slip — a quick unit check catches it.
§7 — Bearings & navigation SL 3.3
Bearings are measured clockwise from North and always written as three figures, $000^\circ$ to $360^\circ$. The whole trick is to turn a journey into a triangle, then reach for the sine or cosine rule.
Back-bearing:the reverse direction is bearing $\pm 180^\circ$ (add if under $180^\circ$, subtract if over)
Interior angle:draw a North line at the turning point; the triangle's interior angle comes from the two bearings there, not from a blind subtraction
TrickSketch a North line at every vertex and mark the bearings onto it. The interior angle of the triangle is rarely just the difference of the two bearings — the back-bearing at the turn is what you actually need.
TrapBearings are always three figures and clockwise ($072^\circ$, not $72^\circ$ or an anticlockwise angle). Once the triangle is drawn it is usually the cosine rule for the distance, then the sine rule for the returning bearing.
§8 — 3-D trigonometry: angle between a line & a plane SL 3.1
The angle between a slanted line and a plane is the angle between the line and its shadow (projection) on that plane. In a cuboid, that means finding a flat base diagonal first, then a single right-angled triangle.
The space diagonal (solid gold) and its shadow, the base diagonal (dashed gold). The angle θ sits between them at the front-bottom corner; the height h is perpendicular to the base.
Base diagonal:$d=\sqrt{l^2+w^2}$ — the flat diagonal across the base (found by Pythagoras)
Line–plane angle:$\theta=\tan^{-1}\!\left(\dfrac{h}{\sqrt{l^2+w^2}}\right)$ — height over base diagonal
Space diagonal:$\sqrt{l^2+w^2+h^2}$ — the full corner-to-corner length
TrickFind the flat base diagonal first — that diagonal is the "adjacent", and the vertical height is the "opposite". The messy 3-D picture then collapses to one ordinary right-angled triangle you can solve with $\tan^{-1}$.
TrapThe angle is measured to the line's projection on the plane, not to a vertical edge or to the nearest side. Identify the right-angled triangle (height $\perp$ base) before reaching for a ratio, or you will use the wrong "adjacent".
A favourite HL crossover marries Trigonometry with Sequences & Series. A triangle built from two sides $a,b$ and their included angle $C$ has area $\tfrac12 ab\sin C$; now nest a chain of similar copies inside it.
Scaling:if every length is multiplied by a factor $k$, each area is multiplied by $k^2$ — not $k$
Geometric areas:the areas form a geometric sequence with common ratio $r=k^2$
Total of $n$:$S_n=A_1\dfrac{1-r^{\,n}}{1-r}$, and if $|k|<1$ the infinite total converges to $S_\infty=\dfrac{A_1}{1-r}$
TrapLengths scale by $k$, but areas scale by $k^2$. Use $r=k^2$ as the common ratio — putting $k$ itself into the series sum is the standard error and collapses the whole answer.
NoteThe same idea powers fractal-style AI models: repeatedly shrinking a shape by a linear factor makes the added areas a convergent geometric series, so a finite total area sits under infinitely many pieces.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
Right-angled triangle, find a side or angle
Label opp/adj/hyp, pick SOH-CAH-TOA; inverse ratio for an angle
Hyp faces the right angle; opp/adj swap with the chosen angle; degree mode
$C$ must sit between $a$ and $b$; never drop the $\tfrac12$
Arc length or sector area
$\dfrac{\theta}{360}2\pi r$ or $\dfrac{\theta}{360}\pi r^2$
Don't swap the two; radians use $r\theta$, $\tfrac12 r^2\theta$
Bearing / navigation
Sketch North lines; cosine rule for distance, sine rule for the bearing
Three figures, clockwise from North; use back-bearings
3-D angle between a line and a plane
Base diagonal $\sqrt{l^2+w^2}$, then $\tan^{-1}\!\big(h/\sqrt{l^2+w^2}\big)$
Project onto the plane first; find the right-angled triangle
Worked Example — AI HL-Style Bearings & Navigation
Question (AI HL Paper 2 style — 7 marks)
A ship sails from port $P$ on a bearing of $060^\circ$ for $8$ km to a buoy $Q$. It then changes course and sails on a bearing of $130^\circ$ for a further $6$ km to a lighthouse $R$. (a) Show that the interior angle $P\hat{Q}R = 110^\circ$. (b) Find the direct distance $PR$. (c) Find the bearing of $R$ from $P$.
Solution
(a) Draw a North line at $Q$. The bearing of $P$ from $Q$ is the back-bearing $060^\circ+180^\circ=240^\circ$, and the bearing of $R$ from $Q$ is $130^\circ$. The interior angle is the difference: $P\hat{Q}R=240^\circ-130^\circ=110^\circ$. (M1)(A1)
(b) Two sides ($PQ=8$, $QR=6$) and the included angle $110^\circ$ — use the cosine rule: $$PR^2=8^2+6^2-2(8)(6)\cos110^\circ=64+36-96(-0.3420)=132.83.$$ So $PR=\sqrt{132.83}=11.5$ km (3 s.f.). (M1)(A1)
(c) Find the angle at $P$ with the sine rule: $\dfrac{\sin(Q\hat{P}R)}{6}=\dfrac{\sin110^\circ}{11.525}$, so $\sin(Q\hat{P}R)=\dfrac{6\sin110^\circ}{11.525}=0.4892$ and $Q\hat{P}R=29.3^\circ$. (M1)(A1) The ship's outbound leg was on bearing $060^\circ$, and $R$ lies a further $29.3^\circ$ clockwise, so the bearing of $R$ from $P$ is $060^\circ+29.3^\circ=089^\circ$ (to the nearest degree). (A1)
Examiner's note: the key move in (a) is the back-bearing at $Q$ — a common error is to write $130^\circ-60^\circ=70^\circ$ for the interior angle, which is wrong. Keep full calculator values through (b) into (c) rather than the rounded $11.5$, then give the final bearing as three figures ($089^\circ$). Carrying the rounded distance can shift the angle by a tenth of a degree.
Common Student Questions
How do I know whether to use the sine rule or the cosine rule?
Count what you are given. If you have a side together with the angle opposite it (plus one more angle or side), use the sine rule. If you have two sides and the angle between them (to find the third side) or all three sides (to find an angle), use the cosine rule. A quick rule of thumb: no matching side–angle pair means the sine rule cannot start, so reach for the cosine rule.
When does the "ambiguous case" actually give two answers?
Only when you use the sine rule to find an angle from two sides and a non-included angle. Because $\sin\theta=\sin(180^\circ-\theta)$, the calculator's acute answer has an obtuse partner. Compute $180^\circ$ minus your angle and check whether that second triangle still has angles summing to under $180^\circ$; if it does, both are valid. Finding a side, or using the cosine rule for an angle, is never ambiguous.
Do I work in degrees or radians for arcs and sectors in AI?
AI runs in degrees by default, so the standard formulas are $\ell=\tfrac{\theta}{360}\times2\pi r$ and $A=\tfrac{\theta}{360}\times\pi r^2$ — the sector as a fraction of the whole circle. If a question gives the angle in radians, switch to the tidier $\ell=r\theta$ and $A=\tfrac12 r^2\theta$, which need no fraction. Match the formula to the units, and make sure your GDC's angle mode agrees.
Why isn't the bearing angle just the difference of the two bearings?
Because the interior angle of your triangle sits at the turning point, and the two directions there are the back-bearing of where you came from and the bearing of where you are going. You almost always have to add or subtract $180^\circ$ to reverse one leg first. Sketch a North line at the vertex, mark both bearings on it, and read the angle between them — a blind subtraction of the two original bearings is the most common bearings mistake.
For the 3-D angle between a line and a plane, what is the "adjacent" side?
It is the line's projection (shadow) on the plane, not any edge of the solid. For a cuboid's space diagonal, first find the flat base diagonal $\sqrt{l^2+w^2}$ — that is the adjacent — while the vertical height $h$ is the opposite. The angle is then $\tan^{-1}\!\big(h/\sqrt{l^2+w^2}\big)$. Getting the projection right is the whole skill; once you have the right-angled triangle it is ordinary SOHCAHTOA.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every right-angle and non-right-angle method, the ambiguous case and 3-D technique in full, with worked edge cases and the exact GDC steps.
Tutorial booklet — 25+ AI HL-style questions from SOHCAHTOA fluency to full bearings, sector and 3-D applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every trigonometry method.
Predicted-paper questions — the exact triangle, bearings and 3-D question types most likely in the next session.
