3.2 Trigonometry in triangles · IB Maths AA

Key terms introduced here

Each is in the Maths AA glossary, with every other term in the subject.

Cosine rule: In any triangle $ABC$ , $c^2 = a^2 + b^2 - 2ab\cos C$ . It links all three sides with one angle and extends Pythagoras's theorem to triangles without a right angle.
Hypotenuse: The longest side of a right-angled triangle, opposite the right angle.
Included angle: The angle between two given sides of a triangle, at the vertex where those two sides meet.
Pythagoras's theorem: In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: $c^2 = a^2 + b^2$ .
Sine rule: In any triangle $ABC$ , $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ , where side $a$ is opposite angle $A$ , and so on. It is used when a side and the angle opposite it are both known.
Trigonometric ratio: For an acute angle $\theta$ in a right-angled triangle, one of $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ . Each depends only on $\theta$ , not on the size of the triangle.

Notes on 3.2 Trigonometry in triangles

24 min readLast reviewed 14 Oct 2026

Trigonometry is the measurement of triangles: it links the lengths of the sides of a triangle to the sizes of its angles. With it, a surveyor can find the width of a river without crossing it, and an engineer can find the length of a roof beam from the pitch of the roof. Any polygon can be split into triangles, so the same tools solve problems about quadrilaterals, polygons and solids.

The work starts with right-angled triangles, where three ratios of sides do everything. Two rules, the sine rule and the cosine rule, then extend trigonometry to triangles with no right angle, and a third formula gives the area of any triangle from two sides and the angle between them.

Right-angled triangles

Naming the sides

In a right-angled triangle, the side opposite the right angle has a special name.

The other two sides are named relative to the angle being used, usually called $\theta$ . The opposite side is across the triangle from $\theta$ , and the adjacent side is next to $\theta$ but is not the hypotenuse.

Right-angled triangle with the right angle at the bottom right and the angle θ at the bottom left. The sloping side is labelled hypotenuse, the vertical side opposite θ is labelled opposite, and the horizontal side next to θ is labelled adjacent. — The hypotenuse is fixed by the right angle; the opposite and adjacent sides depend on which angle is θ.

If the other acute angle is used instead, the opposite and adjacent sides swap, while the hypotenuse stays the same. Labelling the three sides relative to the angle in question is the first step of every right-angled calculation.

Pythagoras's theorem

When two sides of a right-angled triangle are known, the third follows from Pythagoras's theorem, without any angles.

To find the hypotenuse, add the squares; to find a shorter side, subtract: $a^2 = c^2 - b^2$ . For example, a triangle with hypotenuse $13$ and one side $5$ has third side $\sqrt{169 - 25} = \sqrt{144} = 12$ .

The trigonometric ratios

All right-angled triangles with the same acute angle $\theta$ are similar: one is an enlargement of another. Enlarging a triangle multiplies every side by the same scale factor, so the ratio of any two sides does not change. Each ratio of sides therefore depends only on the angle $\theta$ , and is given a name.

The trigonometric ratios are read as "sine", "cosine" and "tangent". Since the hypotenuse is the longest side, $\sin\theta$ and $\cos\theta$ are always between $0$ and $1$ for an acute angle, while $\tan\theta$ can take any positive value. Dividing the first ratio by the second gives $\frac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}} = \tan\theta$ , a relationship used throughout the topic.

Finding a side

To find a side, choose the ratio that links the known side and the unknown side, substitute the angle, and rearrange. If the unknown is on the top of the fraction, multiply; if it is on the bottom, the unknown and the ratio swap places.

When the unknown side is the hypotenuse, or when the known side is the hypotenuse but the unknown is adjacent, the same method applies with a different ratio. The next example has the unknown on the bottom of the fraction.

In a right-angled triangle, the side adjacent to an angle of $28°$ is $9$ m. Find (a) the hypotenuse and (b) the side opposite the $28°$ angle.

Show the solution

(a) Adjacent and hypotenuse: use cosine.

\cos 28° = \frac{9}{h} \quad \Rightarrow \quad h = \frac{9}{\cos 28°} = 10.193\ldots \approx 10.2 \text{ m}

The unknown was on the bottom, so it ends up dividing: $h = \frac{9}{\cos 28°}$ , not $9\cos 28°$ .

(b) Opposite and adjacent: use tangent.

\tan 28° = \frac{y}{9} \quad \Rightarrow \quad y = 9\tan 28° = 4.785\ldots \approx 4.79 \text{ m}

Check with Pythagoras's theorem: $\sqrt{9^2 + 4.785^2} = 10.19$ , which agrees with (a).

Finding an angle

To find an angle from two sides, find the value of the ratio and then use the inverse function on a calculator: $\sin^{-1}$ , $\cos^{-1}$ or $\tan^{-1}$ (also written arcsin, arccos and arctan). The inverse function answers the question "which angle has this sine (or cosine, or tangent)?".

Isosceles triangles

An isosceles triangle has no right angle, but the line from its apex to the midpoint of its base is a line of symmetry and meets the base at a right angle. It splits the triangle into two identical right-angled triangles, so right-angled trigonometry still applies.

An isosceles triangle has two equal sides of $12$ cm and a base of $10$ cm. Find (a) the base angles, (b) the angle at the apex and (c) the area of the triangle.

Show the solution

(a) The line of symmetry cuts the base into two halves of $5$ cm. In each right-angled triangle, the hypotenuse is $12$ cm and the side adjacent to the base angle $\theta$ is $5$ cm.

\cos\theta = \frac{5}{12} \quad \Rightarrow \quad \theta = 65.375\ldots° \approx 65.4°

(b) The angles of a triangle add to $180°$ .

\text{Apex angle} = 180° - 2 \times 65.375\ldots° = 49.249\ldots° \approx 49.2°

(c) The height is the third side of the right-angled triangle.

h = \sqrt{12^2 - 5^2} = \sqrt{119}, \qquad \text{Area} = \frac{1}{2} \times 10 \times \sqrt{119} = 5\sqrt{119} \approx 54.5 \text{ cm}^2

Triangles without a right angle

The labelling convention

For a general triangle, the vertices are labelled with capital letters $A$ , $B$ , $C$ , and the same capital letter stands for the angle at that vertex. Each side is labelled with the lower-case letter of the opposite vertex: side $a$ is opposite angle $A$ , side $b$ is opposite angle $B$ , and side $c$ is opposite angle $C$ .

Triangle ABC with angles marked at A, B and C. Side a = BC is opposite A, side b = AC is opposite B, and side c = AB is opposite C. — Each side takes the lower-case letter of the angle opposite it: a faces A, b faces B and c faces C.

The convention makes the formulas easy to apply to triangles with other letters. In a triangle $PQR$ , the side opposite $P$ is $QR$ , and it plays the role of $p$ . The angle between two given sides, at the vertex where they meet, is their included angle: in triangle $ABC$ , angle $C$ is the included angle of sides $a$ and $b$ .

An altitude splits a triangle into two right-angled triangles

All three results for general triangles come from one construction. Drop a perpendicular from $C$ to the side $AB$ , meeting it at $D$ . This altitude $CD$ , of length $h$ , splits triangle $ABC$ into two right-angled triangles, $ADC$ and $BDC$ . Let $AD = x$ , so that $DB = c - x$ .

Triangle ABC with base AB. The altitude from C meets AB at D at a right angle and has length h. AD = x and DB = c − x. Side b = AC and side a = BC. — The altitude CD creates two right-angled triangles, ADC and BDC, which share the height h.

In triangle $ADC$ , the hypotenuse is $b$ , so

h = b\sin A \qquad \text{and} \qquad x = b\cos A

In triangle $BDC$ , the hypotenuse is $a$ , so

h = a\sin B

These three facts give the sine rule, the cosine rule and the area formula.

The sine rule

Both expressions for the altitude equal $h$ , so $b\sin A = a\sin B$ . Dividing both sides by $\sin A\sin B$ gives

\frac{a}{\sin A} = \frac{b}{\sin B}

Dropping the altitude from $A$ instead gives $\frac{b}{\sin B} = \frac{c}{\sin C}$ in the same way, so all three ratios are equal. When one angle is obtuse, the foot of an altitude can fall outside the triangle; the same result still holds, because $\sin(180° - \theta) = \sin\theta$ , a fact explained with the unit circle in 3.5.

The sine rule links each side to the angle opposite it. It is used when a side and its opposite angle are both known, together with one more side or angle. Only two of the three fractions are ever used at once. Put the unknown on the top: the first form to find a side, the upside-down form to find an angle.

Finding a side with the sine rule

When two angles are known, the third is found from the angle sum, $180°$ . Any known side then has a known opposite angle, and the sine rule gives every other side.

In triangle $ABC$ , $A = 40°$ , $B = 65°$ and $a = 12$ cm. Find $b$ and $c$ .

Show the solution

Use the pair $a$ and $A$ , which are both known, with the side $b$ opposite $B$ .

\frac{b}{\sin 65°} = \frac{12}{\sin 40°} \quad \Rightarrow \quad b = \frac{12\sin 65°}{\sin 40°} = 16.919\ldots \approx 16.9 \text{ cm}

The third angle is $C = 180° - 40° - 65° = 75°$ .

\frac{c}{\sin 75°} = \frac{12}{\sin 40°} \quad \Rightarrow \quad c = \frac{12\sin 75°}{\sin 40°} = 18.032\ldots \approx 18.0 \text{ cm}

Check: the largest angle, $C = 75°$ , is opposite the longest side, $c$ .

Finding an angle with the sine rule

To find an angle, two sides and an angle opposite one of them must be known. Use the upside-down form so that the unknown sine is on the top.

In triangle $ABC$ , $a = 9$ cm, $b = 7$ cm and $A = 70°$ . Find angle $B$ and angle $C$ .

Show the solution

The known pair is $a$ and $A$ ; the unknown angle $B$ is opposite the known side $b$ .

\frac{\sin B}{7} = \frac{\sin 70°}{9} \quad \Rightarrow \quad \sin B = \frac{7\sin 70°}{9} = 0.7308\ldots

B = \sin^{-1}(0.7308\ldots) = 46.959\ldots° \approx 47.0°

Since $b < a$ , angle $B$ must be smaller than angle $A = 70°$ , so $B$ is acute and this is the only possibility.

C = 180° - 70° - 46.959\ldots° = 63.040\ldots° \approx 63.0°

The cosine rule

The sine rule cannot start if no side is known together with its opposite angle. This happens with two sides and the included angle, or with three sides and no angles. The cosine rule handles both cases.

Proving the cosine rule

Return to the altitude construction. In triangle $BDC$ , Pythagoras's theorem gives $a^2 = h^2 + (c - x)^2$ . In triangle $ADC$ , $h^2 = b^2 - x^2$ and $x = b\cos A$ . Substituting:

\begin{aligned} a^2 &= (b^2 - x^2) + (c^2 - 2cx + x^2) \\ &= b^2 + c^2 - 2cx \\ &= b^2 + c^2 - 2bc\cos A \end{aligned}

Relabelling the triangle gives the same result for any side; the form with $c^2$ on the left is the standard one.

The cosine rule contains all three sides and one angle, the angle opposite the side on the left. When $C = 90°$ , $\cos C = 0$ and the rule becomes $c^2 = a^2 + b^2$ , so the cosine rule is Pythagoras's theorem with a correction term for triangles that are not right-angled. If $C$ is obtuse, $\cos C$ is negative, the correction term is added, and $c$ is longer than it would be in a right-angled triangle.

Finding a side: two sides and the included angle

Finding an angle: three sides

With three sides known, the rearranged form gives the cosine of any angle. The angle to find is the one opposite the side that has the minus sign in front of it.

A triangle has sides $7$ cm, $8$ cm and $13$ cm. Find the largest angle.

Show the solution

The largest angle is opposite the longest side, $13$ cm. Call this angle $C$ , with $a = 7$ , $b = 8$ and $c = 13$ .

\cos C = \frac{7^2 + 8^2 - 13^2}{2 \times 7 \times 8} = \frac{49 + 64 - 169}{112} = \frac{-56}{112} = -\frac{1}{2}

C = \cos^{-1}\left(-\frac{1}{2}\right) = 120°

The cosine is negative, so the angle is obtuse. Unlike the sine rule, the cosine rule never gives two possible angles: each value of $\cos C$ between $-1$ and $1$ matches exactly one angle between $0°$ and $180°$ .

The cosine rule also works with exact values. Using $\cos 60° = \frac{1}{2}$ (one of the exact values derived in 3.5), a triangle with $a = 5$ , $b = 8$ and $C = 60°$ has

c^2 = 25 + 64 - 2 \times 5 \times 8 \times \frac{1}{2} = 89 - 40 = 49 \quad \Rightarrow \quad c = 7

The area of a triangle

The area of a triangle is half the base times the perpendicular height. In the altitude construction, the base is $c$ and the height is $h = b\sin A$ , so

\text{Area} = \frac{1}{2} \times c \times b\sin A = \frac{1}{2}bc\sin A

Relabelling gives the same pattern for any pair of sides: half the product of two sides times the sine of their included angle.

Finding an angle from the area

If the area and two sides are known, the formula gives the sine of the included angle. Two angles between $0°$ and $180°$ have the same sine, $\theta$ and $180° - \theta$ , so there are usually two triangles with the given area.

A triangle has sides $8$ cm and $10$ cm, and area $30 \text{ cm}^2$ . Find the possible sizes of the angle between these two sides.

Show the solution

Substitute into the area formula.

\frac{1}{2} \times 8 \times 10 \times \sin\theta = 30 \quad \Rightarrow \quad \sin\theta = \frac{30}{40} = 0.75

The calculator gives the acute angle.

\theta = \sin^{-1}(0.75) = 48.590\ldots° \approx 48.6°

The obtuse angle $180° - 48.590\ldots° = 131.409\ldots° \approx 131.4°$ has the same sine, and also gives a valid triangle. Both answers are correct.

Both triangles have the same base and the same perpendicular height, so the same area.

Choosing the right method

A triangle is fixed by three pieces of information, at least one of which is a side. Which rule to use depends only on which three are known.

Information given	To find	Use
A right angle and two sides	The third side	Pythagoras's theorem
A right angle, one side and one acute angle (or two sides)	A side or an angle	$\sin$ , $\cos$ , $\tan$
Two angles and any side	Another side	Sine rule (third angle from $180°$ )
Two sides and an angle opposite one of them	Another angle	Sine rule (watch for a second solution)
Two sides and the included angle	The third side	Cosine rule
Three sides	Any angle	Cosine rule, rearranged
Two sides and the included angle	The area	$\frac{1}{2}ab\sin C$

A quick test separates the sine rule from the cosine rule: if a side and the angle opposite it are both known, the sine rule can be used; if not, start with the cosine rule. Many problems use both, one after the other.

In the quadrilateral $ABCD$ , $AB = 6$ cm, $BC = 9$ cm, angle $ABC = 110°$ , $AD = 8$ cm and angle $CAD = 35°$ .

(a) Find the length of the diagonal $AC$ .

(b) Find the area of $ABCD$ .

(c) Find the length of $CD$ .

Show the solution

The diagonal AC splits the quadrilateral into two triangles that can each be solved.

(a) In triangle $ABC$ , two sides and the included angle are known: cosine rule.

AC^2 = 6^2 + 9^2 - 2 \times 6 \times 9 \times \cos 110° = 117 + 36.94\ldots = 153.94\ldots

AC = 12.407\ldots \approx 12.4 \text{ cm}

Since $\cos 110°$ is negative, subtracting it adds to $117$ .

(b) Add the areas of the two triangles, using the unrounded $AC$ .

\text{Area of } ABC = \frac{1}{2} \times 6 \times 9 \times \sin 110° = 25.371\ldots

\text{Area of } ACD = \frac{1}{2} \times 12.407\ldots \times 8 \times \sin 35° = 28.465\ldots

\text{Area of } ABCD = 53.837\ldots \approx 53.8 \text{ cm}^2

(c) In triangle $ACD$ , two sides ( $AC$ and $AD$ ) and their included angle are known: cosine rule again.

CD^2 = 12.407\ldots^2 + 8^2 - 2 \times 12.407\ldots \times 8 \times \cos 35° = 55.32\ldots

CD = 7.438\ldots \approx 7.44 \text{ cm}

Exam technique for triangle problems

(GDC allowed.) In triangle $PQR$ , $PQ = 10$ cm, $PR = 7$ cm and $QR = 14$ cm.

(a) Find the size of angle $QPR$ . [3 marks]

(b) Find the area of triangle $PQR$ . [2 marks]

(c) Find the shortest distance from $P$ to the side $QR$ . [2 marks]

(d) Find the size of angle $PQR$ . [2 marks]

Show the solution

(a) Three sides are known, so use the cosine rule. The angle at $P$ is opposite $QR = 14$ .

\cos(Q\hat{P}R) = \frac{10^2 + 7^2 - 14^2}{2 \times 10 \times 7} = \frac{-47}{140}

Q\hat{P}R = 109.616\ldots° \approx 109.6°

(b) Sides $PQ$ and $PR$ have included angle $Q\hat{P}R$ .

\text{Area} = \frac{1}{2} \times 10 \times 7 \times \sin(109.616\ldots°) = 32.968\ldots \approx 33.0 \text{ cm}^2

(c) The shortest distance from $P$ to $QR$ is the perpendicular height $d$ onto the base $QR$ . Use $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ .

\frac{1}{2} \times 14 \times d = 32.968\ldots \quad \Rightarrow \quad d = 4.709\ldots \approx 4.71 \text{ cm}

(d) Now a side and its opposite angle are known ( $QR$ and $Q\hat{P}R$ ), so the sine rule can be used. $P\hat{Q}R$ is opposite $PR = 7$ . Since $7$ is not the longest side, the angle is acute.

\frac{\sin(P\hat{Q}R)}{7} = \frac{\sin(109.616\ldots°)}{14} \quad \Rightarrow \quad \sin(P\hat{Q}R) = 0.47098\ldots

P\hat{Q}R = 28.098\ldots° \approx 28.1°

The next question is set without a calculator. It uses the exact values $\cos 60° = \frac{1}{2}$ and $\sin 60° = \frac{\sqrt{3}}{2}$ , and combines the cosine rule with a quadratic equation.

(No calculator.) In triangle $PQR$ , $PQ = x$ cm, $PR = (x + 2)$ cm, $QR = 2\sqrt{7}$ cm and angle $QPR = 60°$ .

(a) Show that $x^2 + 2x - 24 = 0$ . [3 marks]

(b) Hence find the value of $x$ . [2 marks]

(c) Find the exact area of triangle $PQR$ . [2 marks]

Show the solution

(a) Angle $QPR$ is the included angle of $PQ$ and $PR$ , and $QR$ is opposite it. Apply the cosine rule, showing each step towards the given equation.

(2\sqrt{7})^2 = x^2 + (x + 2)^2 - 2x(x + 2)\cos 60°

28 = x^2 + x^2 + 4x + 4 - 2x(x + 2) \times \frac{1}{2}

28 = 2x^2 + 4x + 4 - x^2 - 2x

28 = x^2 + 2x + 4

x^2 + 2x - 24 = 0

(b) "Hence" means use the equation from (a). Factorise.

(x + 6)(x - 4) = 0 \quad \Rightarrow \quad x = -6 \text{ or } x = 4

A length must be positive, so reject $x = -6$ . Therefore $x = 4$ .

(c) The sides are $PQ = 4$ and $PR = 6$ , with included angle $60°$ .

\text{Area} = \frac{1}{2} \times 4 \times 6 \times \sin 60° = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm}^2

Self-check

To do:
I can label the hypotenuse, opposite and adjacent sides relative to a given angle.
To do:
I can use $\sin$ , $\cos$ and $\tan$ to find a side or an angle in a right-angled triangle.
To do:
I can label any triangle so that side $a$ is opposite angle $A$ , and identify an included angle.
To do:
I can use the sine rule to find a side, or an angle when the side opposite it is known.
To do:
I can use the cosine rule to find a side from two sides and the included angle, and an angle from three sides.
To do:
I can find the area of a triangle with $\frac{1}{2}ab\sin C$ , and find the two possible angles from a given area.
To do:
I can prove the sine rule, the cosine rule and the area formula from an altitude.
To do:
I can decide which rule to use from the information given, and combine rules in multi-step problems.

Questions on 3.2

92 exam-style questions on this subtopic, written in the style of the IB papers. The first has its markscheme open. Counted on 20 October 2026

Question 1 · AA-000617 marks

The cuboid $\mathrm{ABCDEFGH}$ has a horizontal rectangular base $\mathrm{ABCD}$ , with $\mathrm{AB} = 6$ cm and $\mathrm{BC} = 3$ cm. The vertices $\mathrm{E}$ , $\mathrm{F}$ , $\mathrm{G}$ and $\mathrm{H}$ lie vertically above $\mathrm{A}$ , $\mathrm{B}$ , $\mathrm{C}$ and $\mathrm{D}$ respectively, and $\mathrm{AE} = 2$ cm.

(a)
Find the length of the diagonal $[\mathrm{AG}]$ .
2 marks
Markscheme (a)
$\mathrm{AG} = \sqrt{6^2 + 3^2 + 2^2}$ (M1)
$= 7$ (cm) A1
(b)
Let $\theta$ be the angle between $[\mathrm{AG}]$ and the base $\mathrm{ABCD}$ . Find the exact value of $\cos\theta$ .
3 marks
Markscheme (b)
$\theta = \mathrm{G\hat{A}C}$ (may be seen on a diagram) (M1)
$\mathrm{AC} = \sqrt{6^2 + 3^2} = \sqrt{45}$ (A1)
$\cos\theta = \frac{\sqrt{45}}{7} \; \left(= \frac{3\sqrt{5}}{7}\right)$ A1
Note: Award A1FT for $\cos\theta = \frac{\sqrt{45}}{\text{their AG}}$ .
(c)
Find the exact value of $\cos\mathrm{G\hat{A}B}$ .
2 marks
Markscheme (c)
triangle $\mathrm{ABG}$ is right-angled at $\mathrm{B}$ (since $[\mathrm{AB}]$ is perpendicular to the face $\mathrm{BCGF}$ ) (R1)
$\cos\mathrm{G\hat{A}B} = \frac{6}{7}$ A1
Note: Award (R1)A1 for a correct answer with no reason seen. Accept use of the cosine rule with $\mathrm{BG} = \sqrt{13}$ .

Question 2 · AA-000646 marks

A skate ramp is a solid wedge with six vertices $\mathrm{A}$ , $\mathrm{B}$ , $\mathrm{C}$ , $\mathrm{D}$ , $\mathrm{E}$ and $\mathrm{F}$ . Its base $\mathrm{ABCD}$ is a horizontal rectangle with $\mathrm{AB} = 14.5$ m and $\mathrm{AD} = 4.2$ m. The face $\mathrm{CDEF}$ is a vertical rectangle, with $\mathrm{E}$ vertically above $\mathrm{D}$ , $\mathrm{F}$ vertically above $\mathrm{C}$ and $\mathrm{DE} = 2.3$ m. The face $\mathrm{ABFE}$ is the sloping surface of the ramp.

(a)
Find the length of $[\mathrm{BE}]$ .
3 marks
(b)
Find the angle between $[\mathrm{BE}]$ and the base of the ramp.
3 marks

Answer it, then compare with the markscheme.Check this one with a free account

Question 3 · AA-000657 marks

Consider the points $\mathrm{A}(5, 5, 2)$ , $\mathrm{B}(1, 2, 0)$ and $\mathrm{C}(5, 0, 4)$ .

(a)
Show that triangle $\mathrm{ABC}$ is isosceles.
3 marks
(b)
Write down the coordinates of $\mathrm{M}$ , the midpoint of $[\mathrm{BC}]$ .
1 mark
(c)
Hence find the exact area of triangle $\mathrm{ABC}$ .
3 marks

Answer it, then compare with the markscheme.Check this one with a free account

Try a quick-fire question

One of 34 short multiple-choice questions on 3.2. Pick an answer to see why it's right or wrong.

Which expression gives the length $AB$ in the triangle shown?

Triangle ABC with angle A = 40°, angle B = 75° and BC = 10 cm.

From the formula booklet

The entries of the mathematics: analysis and approaches formula booklet that 3.2 uses. Each opens in the explained booklet.

Key terms introduced here

Notes on 3.2 Trigonometry in triangles

Right-angled triangles

Naming the sides

Pythagoras's theorem

The trigonometric ratios

Finding a side

Finding an angle

Isosceles triangles

Triangles without a right angle

The labelling convention

An altitude splits a triangle into two right-angled triangles

The sine rule

Finding a side with the sine rule

Finding an angle with the sine rule

The cosine rule

Proving the cosine rule

Finding a side: two sides and the included angle

Finding an angle: three sides

The area of a triangle

Finding an angle from the area

Choosing the right method

Exam technique for triangle problems

Self-check

Questions on 3.2

Markscheme (a)

Markscheme (b)

Markscheme (c)

Try a quick-fire question

From the formula booklet

Carry on with 3.2 in Augment