Find a side or an angle

Choose the rule

In 9 of 20 recent exams
2% of the marks
Paper 1 or 2
1 to 5 marks

How it is set in the exam

Paper 2 (calculator) mostly (16 of the 20 parts set since 2021), 1–5 marks a part (the 1-mark part is a step that feeds the next part): a short Section A question (a side, then an angle), or the opening parts of a longer question whose later parts reuse the side. On Paper 1 (no calculator) (4 parts) it is exact: a special angle or a cosine given as a fraction, often with sides in terms of $x$ . Common (9 of 20 exam versions), and it is the solving step inside every other type in this guide.

Spot it in the exam

Exam phrases to spot (4)

“find the length of $[BC]$ ”
“find the size of angle $ACB$ ”
“calculate $PR$ ”
“show that $BC^2 = \ldots$ ”

In disguise (4)

a triangle drawn inside a 3D figure whose three sides are already given as lengths (no coordinates to use): it is a flat triangle like any other
an angle that grows at a constant rate: it is rate × time, an expression in $t$ (angular speed), and the sine rule then gives a length in terms of $t$
a condition to turn into a fact first, such as a line that splits a region into two parts of equal area, which fixes a point before any rule applies
a slope given by its gradient, not by an angle (a roof, a ramp): the gradient is the $\tan$ of the angle with the horizontal (The angle of a line), so $\tan^{-1}$ of it gives that angle, and the angle with a vertical is $90°$ minus it

Variants (4)

SAS for the third side, or SSS for an angle, then the sine rule for another angle (one value: in a fixed triangle only the largest angle can be obtuse, so the sine rule is used for a smaller angle and the cosine rule for the largest)
two triangles sharing a side: right-angled trig in one, then the sine or cosine rule in the other, sometimes with an angle made of two pieces
an angle found in a small right triangle and reused in a larger right triangle of the same structure
exact on : a special angle or a fractional cosine; sides in terms of $x$ give a quadratic with a root to reject

Looks like it, but isn't (7)

directions from north: Bearings
a line of sight from an observer (elevation or depression), or angles marked from the horizontal at the top of a building or mast, to be turned into angles of the triangle: Angles of elevation and depression
two sides and a non-included angle that faces the shorter side: The ambiguous case
an area asked for, or given as a number: Area of a triangle (a condition such as two parts of equal area, which only fixes a point, stays here)
a triangle made of two radii, or a chord of a circle: Chords, segments and regions
a face or a cut through a solid whose sides must first come from 3D data (coordinates, half a diagonal, equal slant edges): Triangles inside a solid
, when only a cosine, a tangent, $\cos 2\theta$ , or angles $\theta$ and $2\theta$ are known and a ratio is what is asked: Exact values without the angle: from one ratio to another

Worked example

SAS with an obtuse angle, then an angle by the sine rule

Paper 2 (calculator)

Question

In triangle $PQR$ , $PQ = 8.5$ cm, $QR = 14.2$ cm and angle $PQR = 104°$ .
(a) Find $PR$ . [3]
(b) Find angle $QPR$ . [3]

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See it move: Which rule fits what you know

What the question gives

You know $PQ = 8.5$ , $QR = 14.2$ and the angle between them, $\hat{Q} = 104°$ : two sides and the included angle. No side comes with its opposite angle, so the sine rule can't start. Cosine rule: $PR^2 = 8.5^2 + 14.2^2 - 2(8.5)(14.2)\cos 104° = 332.28\ldots$ , so $PR = 18.2$ cm.
$\hat{Q} = 104°$ is obtuse, so $\cos \hat{Q} < 0$ and the last term adds: $PR = 18.2$ is longer than $\sqrt{PQ^2 + QR^2} = 16.5$ . An acute angle takes away instead.
(a) The $104°$ at $Q$ sits between the two known sides, $8.5$ and $14.2$ : SAS, so the cosine rule: $PR^2 = 332.28\ldots$ , $PR = 18.2$ cm. In radian mode the same working reads $22.4$ cm: the method marks stay, the answer goes.
(b) After (a), $PR$ and $\hat{Q}$ are a complete pair: the sine rule with the stored $18.228\ldots$ , not $18.2$ : $\sin \hat{P} = 0.75584\ldots$ , $\hat{P} = 49.1°$ . $\sin^{-1}$ gives the acute angle; the obtuse $130.9°$ can't sit beside $104°$ .
Check: $49.1° + 104° + 26.9° = 180°$ , the largest angle, $\hat{Q}$ , faces the longest side, $PR$ , and the mode matches the question's degree signs.

Which method?

a right angle
Rule
SOH CAH TOA; Pythagoras for the third side
Watch
name the sides from the angle you use
two sides and the angle between them (SAS)
Rule
the cosine rule for the third side
Watch
the angle must be the one between the two sides
three sides (SSS)
Rule
the cosine rule for an angle
Watch
largest angle first (it faces the longest side)
two angles and a side
Rule
the angle sum for the third angle, then the sine rule
Watch
pair each side with its opposite angle
two sides and a non-included angle that is obtuse, or faces the longer known side
Rule
the sine rule: one triangle
Watch
$\sin^{-1}$ gives it; $180° - B$ doesn't fit
two sides and a non-included angle that is acute and faces the shorter known side
Rule
The ambiguous case
Watch
two triangles, one right-angled triangle, or none: compare the side with the height $b\sin A$

What you know	Rule	Watch
a right angle	SOH CAH TOA; Pythagoras for the third side	name the sides from the angle you use
two sides and the angle between them (SAS)	the cosine rule for the third side	the angle must be the one between the two sides
three sides (SSS)	the cosine rule for an angle	largest angle first (it faces the longest side)
two angles and a side	the angle sum for the third angle, then the sine rule	pair each side with its opposite angle
two sides and a non-included angle that is obtuse, or faces the longer known side	the sine rule: one triangle	$\sin^{-1}$ gives it; $180° - B$ doesn't fit
two sides and a non-included angle that is acute and faces the shorter known side	The ambiguous case	two triangles, one right-angled triangle, or none: compare the side with the height $b\sin A$

Dropping a perpendicular to split a triangle into two right-angled triangles is always an alternative, and it scores in full.

Steps

Draw and label
the triangle: sides named after the opposite vertices, every given length and angle marked, the target marked.
List what you know,
and look for a complete pair (a side with its opposite angle).
Choose the rule
from the table. In a chain of triangles, start in the triangle that has enough information, and find the shared side there.
Solve,
keeping unrounded values for the next step (keep full-accuracy values). For an angle from a sine: in a triangle already fixed by SAS or SSS, use the sine rule only for an angle that is not the largest (it is acute, and $\sin^{-1}$ gives it: angle from a ratio) and find the largest with the cosine rule; only with SSA data ask whether $180° - B$ fits (The ambiguous case).
Check:
the angles add up to $180°$ , the longest side faces the largest angle, and the mode matches the question.

Common mistakes

The sine rule with no complete pair
(SAS or SSS): every fraction then has two unknowns. Use the cosine rule.
The wrong angle in the cosine rule
it must be the angle between the two sides you substitute, facing the side you want.
$(a^2 + b^2 - 2ab)\cos C$
typed into the GDC instead of $a^2 + b^2 - 2ab\cos C$ , or the final square root forgotten.

More mistakes (5)

The largest angle by the sine rule
$\sin^{-1}$ never returns an obtuse angle, so an obtuse answer is lost. Find the largest angle with the cosine rule.
$180° - B$ given when it can't fit
(the triangle already has an obtuse angle, or $B$ faces the shorter side), or forgotten when it can (that is The ambiguous case).
The wrong mode
it keeps the method marks and loses the answer mark (check the settings). Example 1.1(a) in radian mode gives $22.4$ cm.
A rounded side carried into the next part
use the stored value; a 3 s.f. side can change the last digit of the next answer.
A rejected root left in
( $x = -10$ in Example 1.3): , and nothing that follows from it scores.

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