Cambridge IGCSE Mathematics 0580 · Extended · Syllabus 2025-2027

Sine Rule Worked ExamplesIGCSE Maths 0580

Exam-style worked examples on the sine rule — finding sides and angles in non-right-angled triangles, deciding when it applies rather than the cosine rule, and the ambiguous case where two triangles satisfy the same data. Diagram-based questions add exact values without a calculator, a quadrilateral split by a diagonal, and a hinged rod with two possible positions.

This page has 11 exam-style IGCSE Maths 0580 sine rule worked examples — 0 Core, 8 Extended and 3 Challenge, 5 with diagrams. Each one shows the full method step by step, the final answer in the form a marker expects, and an examiner tip on where marks are commonly lost.

Example 1ExtendedFinding a side[3 marks]

Finding a side with the sine rule

In triangle $ABC$ , $A = 40^{\circ}$ , $B = 75^{\circ}$ and side $a = 10\text{ cm}$ .

Calculate the length of side $b$ , correct to $3$ significant figures.

Example 2ExtendedFinding a side[4 marks]

Finding the third side

Using the same triangle ( $A = 40^{\circ}$ , $B = 75^{\circ}$ , $a = 10\text{ cm}$ ), calculate side $c$ .

Example 3ExtendedFinding an angle[3 marks]

Finding an angle with the sine rule

In triangle $ABC$ , $a = 8\text{ cm}$ , $b = 11\text{ cm}$ and $A = 35^{\circ}$ .

Calculate the acute value of angle $B$ , correct to $1$ decimal place.

Example 4ChallengeAmbiguous case[4 marks]

The ambiguous case

Using the same data ( $a = 8\text{ cm}$ , $b = 11\text{ cm}$ , $A = 35^{\circ}$ ), show that there is a second possible triangle and find the obtuse value of angle $B$ .

Example 5ExtendedChoosing a method[3 marks]

Choosing between the sine and cosine rules

For each triangle, state whether the sine rule or the cosine rule should be used, and why.

(a) Two angles and one side are known. (b) Three sides are known and an angle is required. (c) Two sides and the angle between them are known.

Example 6ExtendedFinding a side[4 marks]

Sine rule in context

Two coastguard stations $A$ and $B$ are $10\text{ km}$ apart. A boat $C$ is sighted such that angle $CAB = 40^{\circ}$ and angle $ABC = 75^{\circ}$ .

Calculate the distance of the boat from station $A$ , correct to $3$ significant figures.

Example 7ExtendedNon-calculator: exact values in the sine rule[4 marks]

Sine rule with exact values

In triangle $ABC$ , angle $BAC = 45^{\circ}$ , angle $ABC = 30^{\circ}$ and $AC = 6\text{ cm}$ .

(a) Without using a calculator, show that $BC = 6\sqrt{2}\text{ cm}$ .

(b) Without calculating any lengths, explain which side of the triangle is the longest.

NOT TO SCALE

Example 8ExtendedQuadrilateral split into triangles[7 marks]

Sine rule twice across a quadrilateral

The diagram shows a quadrilateral $ABCD$ with diagonal $BD$ . $AB = 9\text{ cm}$ , $CD = 7.5\text{ cm}$ , angle $DAB = 72^{\circ}$ , angle $ABD = 48^{\circ}$ and angle $BCD = 115^{\circ}$ .

(a) Calculate $BD$ .

(b) Calculate angle $CBD$ .

(c) Calculate angle $ABC$ .

NOT TO SCALE

1
(a) In triangle $ABD$ , find the third angle so that the known side $AB$ has a matching opposite angle.
$\angle ADB = 180^{\circ} - 72^{\circ} - 48^{\circ} = 60^{\circ}$
2
$BD$ is opposite the $72^{\circ}$ angle and $AB = 9$ is opposite the $60^{\circ}$ angle.
$\frac{BD}{\sin 72^{\circ}} = \frac{9}{\sin 60^{\circ}} \quad \Rightarrow \quad BD = \frac{9\sin 72^{\circ}}{\sin 60^{\circ}} = 9.8836\ldots$
3
(b) In triangle $BCD$ , $BD$ is opposite the $115^{\circ}$ angle, which makes a matching pair. $CD = 7.5$ is opposite angle $CBD$ . Write the sine rule with the angles on top.
$\frac{\sin(\angle CBD)}{7.5} = \frac{\sin 115^{\circ}}{9.8836\ldots}$
4
Rearrange and use the unrounded $BD$ .
$\sin(\angle CBD) = \frac{7.5\sin 115^{\circ}}{9.8836\ldots} = 0.68773\ldots \quad \Rightarrow \quad \angle CBD = 43.450\ldots^{\circ}$
5
Check for a second value. The obtuse option $180^{\circ} - 43.45^{\circ} = 136.55^{\circ}$ cannot be right, because the triangle already contains $115^{\circ}$ .
6
(c) Angle $ABC$ is made of the two angles at $B$ on either side of the diagonal.
$\angle ABC = 48^{\circ} + 43.450\ldots^{\circ} = 91.450\ldots^{\circ}$

Answer

(a) $9.88\text{ cm}$ (b) $43.5^{\circ}$ (c) $91.5^{\circ}$

Examiner tip

Split the quadrilateral at the diagonal and treat each triangle separately. The diagonal is the link between them, so carry its full calculator value into part (b). The March 2026 examiner report found that "Many candidates did not keep a value to at least 4 significant figures throughout their working". A common error in (b) is to pair $7.5$ with the $115^{\circ}$ angle. $7.5$ is opposite angle $CBD$ , and $BD$ is the side opposite $115^{\circ}$ . The ambiguous case cannot arise here: a triangle that already has an obtuse angle cannot have a second one.

Example 9ChallengeAmbiguous case[7 marks]

Two positions of a hinged rod

A rod $BC$ is hinged at $B$ . The end $C$ must rest on a straight rail that starts at $A$ , where $AB = 12\text{ cm}$ and the rail makes an angle of $32^{\circ}$ with $AB$ . When $BC = 8\text{ cm}$ , there are two possible positions, $C_1$ and $C_2$ , as shown in the diagram.

(a) Calculate the shortest length the rod could have and still reach the rail.

(b) For $BC = 8\text{ cm}$ , use the sine rule to find angle $AC_1B$ and angle $AC_2B$ .

(c) Calculate the distance $C_1C_2$ between the two positions.

NOT TO SCALE

1
(a) The shortest rod reaches the rail at a right angle. That is the perpendicular distance from $B$ to the rail, opposite the $32^{\circ}$ angle in a right-angled triangle with hypotenuse $AB = 12$ .
$12\sin 32^{\circ} = 6.3590\ldots$
2
(b) In triangle $ABC$ , $BC = 8$ is opposite the $32^{\circ}$ angle, and $AB = 12$ is opposite angle $ACB$ .
$\frac{\sin(\angle ACB)}{12} = \frac{\sin 32^{\circ}}{8} \quad \Rightarrow \quad \sin(\angle ACB) = \frac{12\sin 32^{\circ}}{8} = 0.79487\ldots$
3
The calculator gives the acute angle. This is the angle at $C_1$ , the position further from $A$ .
$\angle AC_1B = \sin^{-1}(0.79487\ldots) = 52.643\ldots^{\circ}$
4
The other angle with the same sine is its supplement. It is valid because $32^{\circ} + 127.36^{\circ} < 180^{\circ}$ .
$\angle AC_2B = 180^{\circ} - 52.643\ldots^{\circ} = 127.356\ldots^{\circ}$
5
(c) $BC_1 = BC_2 = 8$ , so triangle $BC_1C_2$ is isosceles. Its base angles are $52.64^{\circ}$ : at $C_2$ this is $180^{\circ} - 127.36^{\circ}$ , the angle on a straight line. Drop the perpendicular from $B$ to the rail. It bisects $C_1C_2$ .
$C_1C_2 = 2 \times 8\cos 52.643\ldots^{\circ} = 9.7082\ldots$

Answer

(a) $6.36\text{ cm}$ (b) $\angle AC_1B = 52.6^{\circ}$ , $\angle AC_2B = 127.4^{\circ}$ (c) $9.71\text{ cm}$

Examiner tip

This is the ambiguous case in physical form. A rod longer than $6.36\text{ cm}$ but shorter than $AB = 12\text{ cm}$ swings onto the rail in two places, and the two angles at $C$ always add up to $180^{\circ}$ . Giving only the calculator's acute value in (b) loses the marks for the second triangle. The March 2025 examiner report found that many candidates used the sine rule but "often used the acute angle" when the obtuse angle was needed. You can check (c) by finding both lengths with the sine rule: $AC_1 = 15.03$ and $AC_2 = 5.32$ , which differ by $9.71$ . ✓

Example 10ExtendedFinding an angle[6 marks]

Show that an angle is correct to 1 decimal place

Three survey pegs, $P$ , $Q$ and $R$ , mark the corners of a triangular plot. Side $PQ$ is $520\text{ m}$ long, side $QR$ is $380\text{ m}$ long and the angle at $P$ is $41^{\circ}$ . The angle at $R$ is acute.

(a) Use the sine rule to show that the angle at $Q$ is $75.1^{\circ}$ , correct to $1$ decimal place.

(b) Work out the area of the plot.

NOT TO SCALE

1
(a) $QR = 380$ is opposite the $41^{\circ}$ angle at $P$ , which gives a matching pair. You cannot reach angle $PQR$ directly, so first find angle $PRQ$ , which is opposite $PQ = 520$ . Put the angles on top.
$\frac{\sin(\angle PRQ)}{520} = \frac{\sin 41^{\circ}}{380} \quad \Rightarrow \quad \sin(\angle PRQ) = \frac{520\sin 41^{\circ}}{380} = 0.89776\ldots$
2
Angle $PRQ$ is acute, so take the calculator value. Keep at least $4$ significant figures.
$\angle PRQ = \sin^{-1}(0.89776\ldots) = 63.865\ldots^{\circ}$
3
Use the angle sum of the triangle, and give more figures than the value you are asked to show.
$\angle PQR = 180^{\circ} - 41^{\circ} - 63.865\ldots^{\circ} = 75.134\ldots^{\circ} = 75.1^{\circ} \text{ (1 d.p.)}$
4
(b) Angle $PQR$ lies between the two known sides $PQ$ and $QR$ , so use $\tfrac{1}{2}ab\sin C$ .
$\text{Area} = \tfrac{1}{2} \times 520 \times 380 \times \sin 75.134\ldots^{\circ} = 95\,493.0\ldots$

Answer

(a) $\sin(\angle PRQ) = 0.89776\ldots$ , $\angle PRQ = 63.865\ldots^{\circ}$ , $\angle PQR = 180 - 41 - 63.865\ldots = 75.134\ldots = 75.1^{\circ}$ (b) $95\,500\text{ m}^2$ (3 s.f.)

Examiner tip

On a question like this, the March 2026 examiner report found that "Many candidates did not keep a value to at least 4 significant figures throughout their working". If you round angle $PRQ$ to $64^{\circ}$ , you get $75^{\circ}$ and cannot show $75.1^{\circ}$ . The same report noted that some candidates used the value they had been asked to show as part of their own working. That proves nothing, so start from the given data. In (b) you may use $75.1^{\circ}$ or your unrounded angle, as both round to $95\,500\text{ m}^2$ . Angle $PRQ$ is stated to be acute because $PQ > QR$ : the obtuse value $116.1^{\circ}$ would give a second, different triangle.

Example 11ChallengeAmbiguous case[7 marks]

A fenced plot with an obtuse angle

A plot of land $EFGH$ is split into two triangular beds by a straight fence $FH$ . Side $EF$ is $6.2\text{ m}$ , side $EH$ is $12.5\text{ m}$ and side $FG$ is $13.8\text{ m}$ . The angle at $E$ is $58^{\circ}$ and the angle at $G$ is $38^{\circ}$ .

(a) Work out the length of the fence $FH$ .

(b) The angle $FHG$ is greater than $90^{\circ}$ . Work out its size.

NOT TO SCALE

1
(a) In triangle $EFH$ the $58^{\circ}$ angle lies between the two known sides, so use the cosine rule.
$FH^2 = 6.2^2 + 12.5^2 - 2(6.2)(12.5)\cos 58^{\circ}$
2
Work out $2 \times 6.2 \times 12.5 \cos 58^{\circ}$ as one term, then subtract it.
$FH^2 = 194.69 - 82.137\ldots = 112.55\ldots \quad \Rightarrow \quad FH = 10.609\ldots$
3
(b) In triangle $FGH$ , $FH$ is opposite the $38^{\circ}$ angle, which gives a matching pair. $FG = 13.8$ is opposite angle $FHG$ . Use the sine rule with the unrounded $FH$ .
$\sin(\angle FHG) = \frac{13.8\sin 38^{\circ}}{10.609\ldots} = 0.80083\ldots$
4
The calculator gives the acute angle with this sine.
$\sin^{-1}(0.80083\ldots) = 53.209\ldots^{\circ}$
5
The angle is obtuse. Sine is positive for obtuse angles too, and $\sin(180^{\circ} - \theta) = \sin\theta$ .
$\angle FHG = 180^{\circ} - 53.209\ldots^{\circ} = 126.790\ldots^{\circ}$
6
Check that the triangle is possible: $38^{\circ} + 126.8^{\circ} = 164.8^{\circ} < 180^{\circ}$ , leaving $15.2^{\circ}$ for angle $GFH$ .

Answer

(a) $10.6\text{ m}$ (b) $126.8^{\circ}$

Examiner tip

On a question of this type, the June 2025 examiner report found that "Only a minority went on from here to correctly give the obtuse value". The sine rule gives the acute angle first, so read the question and the diagram before you stop. Keep $FH$ unrounded: using $FH = 10.6$ in (b) gives $126.7^{\circ}$ and loses the accuracy mark. The same session's report warned that "premature approximation was seen and time was lost unnecessarily". In (a), the June 2025 report also noted that "a very common incorrect answer came from a misconception in calculating" with the cosine rule, so evaluate $2bc\cos A$ on its own first.

These worked examples are written by Tutopiya to cover the same syllabus content as the official Cambridge IGCSE 0580 specification (2025-2027). They are exam-style questions, not reproductions of past paper questions.

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Sine Rule Worked ExamplesIGCSE Maths 0580

Example 1ExtendedFinding a side[3 marks]

Finding a side with the sine rule

In triangle $ABC$ , $A = 40^{\circ}$ , $B = 75^{\circ}$ and side $a = 10\text{ cm}$ .

Calculate the length of side $b$ , correct to $3$ significant figures.

Example 2ExtendedFinding a side[4 marks]

Finding the third side

Using the same triangle ( $A = 40^{\circ}$ , $B = 75^{\circ}$ , $a = 10\text{ cm}$ ), calculate side $c$ .

Example 3ExtendedFinding an angle[3 marks]

Finding an angle with the sine rule

In triangle $ABC$ , $a = 8\text{ cm}$ , $b = 11\text{ cm}$ and $A = 35^{\circ}$ .

Calculate the acute value of angle $B$ , correct to $1$ decimal place.

Example 4ChallengeAmbiguous case[4 marks]

The ambiguous case

Using the same data ( $a = 8\text{ cm}$ , $b = 11\text{ cm}$ , $A = 35^{\circ}$ ), show that there is a second possible triangle and find the obtuse value of angle $B$ .

Example 5ExtendedChoosing a method[3 marks]

Choosing between the sine and cosine rules

For each triangle, state whether the sine rule or the cosine rule should be used, and why.

(a) Two angles and one side are known. (b) Three sides are known and an angle is required. (c) Two sides and the angle between them are known.

Example 6ExtendedFinding a side[4 marks]

Sine rule in context

Two coastguard stations $A$ and $B$ are $10\text{ km}$ apart. A boat $C$ is sighted such that angle $CAB = 40^{\circ}$ and angle $ABC = 75^{\circ}$ .

Calculate the distance of the boat from station $A$ , correct to $3$ significant figures.

Example 7ExtendedNon-calculator: exact values in the sine rule[4 marks]

Sine rule with exact values

In triangle $ABC$ , angle $BAC = 45^{\circ}$ , angle $ABC = 30^{\circ}$ and $AC = 6\text{ cm}$ .

(a) Without using a calculator, show that $BC = 6\sqrt{2}\text{ cm}$ .

(b) Without calculating any lengths, explain which side of the triangle is the longest.

NOT TO SCALE

Example 8ExtendedQuadrilateral split into triangles[7 marks]

Sine rule twice across a quadrilateral

The diagram shows a quadrilateral $ABCD$ with diagonal $BD$ . $AB = 9\text{ cm}$ , $CD = 7.5\text{ cm}$ , angle $DAB = 72^{\circ}$ , angle $ABD = 48^{\circ}$ and angle $BCD = 115^{\circ}$ .

(a) Calculate $BD$ .

(b) Calculate angle $CBD$ .

(c) Calculate angle $ABC$ .

NOT TO SCALE

1
(a) In triangle $ABD$ , find the third angle so that the known side $AB$ has a matching opposite angle.
$\angle ADB = 180^{\circ} - 72^{\circ} - 48^{\circ} = 60^{\circ}$
2
$BD$ is opposite the $72^{\circ}$ angle and $AB = 9$ is opposite the $60^{\circ}$ angle.
$\frac{BD}{\sin 72^{\circ}} = \frac{9}{\sin 60^{\circ}} \quad \Rightarrow \quad BD = \frac{9\sin 72^{\circ}}{\sin 60^{\circ}} = 9.8836\ldots$
3
(b) In triangle $BCD$ , $BD$ is opposite the $115^{\circ}$ angle, which makes a matching pair. $CD = 7.5$ is opposite angle $CBD$ . Write the sine rule with the angles on top.
$\frac{\sin(\angle CBD)}{7.5} = \frac{\sin 115^{\circ}}{9.8836\ldots}$
4
Rearrange and use the unrounded $BD$ .
$\sin(\angle CBD) = \frac{7.5\sin 115^{\circ}}{9.8836\ldots} = 0.68773\ldots \quad \Rightarrow \quad \angle CBD = 43.450\ldots^{\circ}$
5
Check for a second value. The obtuse option $180^{\circ} - 43.45^{\circ} = 136.55^{\circ}$ cannot be right, because the triangle already contains $115^{\circ}$ .
6
(c) Angle $ABC$ is made of the two angles at $B$ on either side of the diagonal.
$\angle ABC = 48^{\circ} + 43.450\ldots^{\circ} = 91.450\ldots^{\circ}$

Answer

(a) $9.88\text{ cm}$ (b) $43.5^{\circ}$ (c) $91.5^{\circ}$

Examiner tip

Example 9ChallengeAmbiguous case[7 marks]

Two positions of a hinged rod

(a) Calculate the shortest length the rod could have and still reach the rail.

(b) For $BC = 8\text{ cm}$ , use the sine rule to find angle $AC_1B$ and angle $AC_2B$ .

(c) Calculate the distance $C_1C_2$ between the two positions.

NOT TO SCALE

1
(a) The shortest rod reaches the rail at a right angle. That is the perpendicular distance from $B$ to the rail, opposite the $32^{\circ}$ angle in a right-angled triangle with hypotenuse $AB = 12$ .
$12\sin 32^{\circ} = 6.3590\ldots$
2
(b) In triangle $ABC$ , $BC = 8$ is opposite the $32^{\circ}$ angle, and $AB = 12$ is opposite angle $ACB$ .
$\frac{\sin(\angle ACB)}{12} = \frac{\sin 32^{\circ}}{8} \quad \Rightarrow \quad \sin(\angle ACB) = \frac{12\sin 32^{\circ}}{8} = 0.79487\ldots$
3
The calculator gives the acute angle. This is the angle at $C_1$ , the position further from $A$ .
$\angle AC_1B = \sin^{-1}(0.79487\ldots) = 52.643\ldots^{\circ}$
4
The other angle with the same sine is its supplement. It is valid because $32^{\circ} + 127.36^{\circ} < 180^{\circ}$ .
$\angle AC_2B = 180^{\circ} - 52.643\ldots^{\circ} = 127.356\ldots^{\circ}$
5
(c) $BC_1 = BC_2 = 8$ , so triangle $BC_1C_2$ is isosceles. Its base angles are $52.64^{\circ}$ : at $C_2$ this is $180^{\circ} - 127.36^{\circ}$ , the angle on a straight line. Drop the perpendicular from $B$ to the rail. It bisects $C_1C_2$ .
$C_1C_2 = 2 \times 8\cos 52.643\ldots^{\circ} = 9.7082\ldots$

Answer

(a) $6.36\text{ cm}$ (b) $\angle AC_1B = 52.6^{\circ}$ , $\angle AC_2B = 127.4^{\circ}$ (c) $9.71\text{ cm}$

Examiner tip

Example 10ExtendedFinding an angle[6 marks]

Show that an angle is correct to 1 decimal place

(a) Use the sine rule to show that the angle at $Q$ is $75.1^{\circ}$ , correct to $1$ decimal place.

(b) Work out the area of the plot.

NOT TO SCALE

1
(a) $QR = 380$ is opposite the $41^{\circ}$ angle at $P$ , which gives a matching pair. You cannot reach angle $PQR$ directly, so first find angle $PRQ$ , which is opposite $PQ = 520$ . Put the angles on top.
$\frac{\sin(\angle PRQ)}{520} = \frac{\sin 41^{\circ}}{380} \quad \Rightarrow \quad \sin(\angle PRQ) = \frac{520\sin 41^{\circ}}{380} = 0.89776\ldots$
2
Angle $PRQ$ is acute, so take the calculator value. Keep at least $4$ significant figures.
$\angle PRQ = \sin^{-1}(0.89776\ldots) = 63.865\ldots^{\circ}$
3
Use the angle sum of the triangle, and give more figures than the value you are asked to show.
$\angle PQR = 180^{\circ} - 41^{\circ} - 63.865\ldots^{\circ} = 75.134\ldots^{\circ} = 75.1^{\circ} \text{ (1 d.p.)}$
4
(b) Angle $PQR$ lies between the two known sides $PQ$ and $QR$ , so use $\tfrac{1}{2}ab\sin C$ .
$\text{Area} = \tfrac{1}{2} \times 520 \times 380 \times \sin 75.134\ldots^{\circ} = 95\,493.0\ldots$

Answer

(a) $\sin(\angle PRQ) = 0.89776\ldots$ , $\angle PRQ = 63.865\ldots^{\circ}$ , $\angle PQR = 180 - 41 - 63.865\ldots = 75.134\ldots = 75.1^{\circ}$ (b) $95\,500\text{ m}^2$ (3 s.f.)

Examiner tip

Example 11ChallengeAmbiguous case[7 marks]

A fenced plot with an obtuse angle

(a) Work out the length of the fence $FH$ .

(b) The angle $FHG$ is greater than $90^{\circ}$ . Work out its size.

NOT TO SCALE

1
(a) In triangle $EFH$ the $58^{\circ}$ angle lies between the two known sides, so use the cosine rule.
$FH^2 = 6.2^2 + 12.5^2 - 2(6.2)(12.5)\cos 58^{\circ}$
2
Work out $2 \times 6.2 \times 12.5 \cos 58^{\circ}$ as one term, then subtract it.
$FH^2 = 194.69 - 82.137\ldots = 112.55\ldots \quad \Rightarrow \quad FH = 10.609\ldots$
3
(b) In triangle $FGH$ , $FH$ is opposite the $38^{\circ}$ angle, which gives a matching pair. $FG = 13.8$ is opposite angle $FHG$ . Use the sine rule with the unrounded $FH$ .
$\sin(\angle FHG) = \frac{13.8\sin 38^{\circ}}{10.609\ldots} = 0.80083\ldots$
4
The calculator gives the acute angle with this sine.
$\sin^{-1}(0.80083\ldots) = 53.209\ldots^{\circ}$
5
The angle is obtuse. Sine is positive for obtuse angles too, and $\sin(180^{\circ} - \theta) = \sin\theta$ .
$\angle FHG = 180^{\circ} - 53.209\ldots^{\circ} = 126.790\ldots^{\circ}$
6
Check that the triangle is possible: $38^{\circ} + 126.8^{\circ} = 164.8^{\circ} < 180^{\circ}$ , leaving $15.2^{\circ}$ for angle $GFH$ .

Answer

(a) $10.6\text{ m}$ (b) $126.8^{\circ}$

Examiner tip

Sine Rule Worked ExamplesIGCSE Maths 0580

Finding a side with the sine rule

Finding the third side

Finding an angle with the sine rule

The ambiguous case

Choosing between the sine and cosine rules

Sine rule in context

Sine rule with exact values

Sine rule twice across a quadrilateral

Two positions of a hinged rod

Show that an angle is correct to 1 decimal place

A fenced plot with an obtuse angle

More IGCSE Maths 0580 worked examples

Trigonometry

Number

Algebra

Functions

Coordinate Geometry

Geometry

Vectors and Transfomations

Mensuration

Sets

Probability

Statistics

Calculator Skills

Sine Rule Worked ExamplesIGCSE Maths 0580

Finding a side with the sine rule

Finding the third side

Finding an angle with the sine rule

The ambiguous case

Choosing between the sine and cosine rules

Sine rule in context

Sine rule with exact values

Sine rule twice across a quadrilateral

Two positions of a hinged rod

Show that an angle is correct to 1 decimal place

A fenced plot with an obtuse angle

More IGCSE Maths 0580 worked examples

Trigonometry

Number

Algebra

Functions

Coordinate Geometry

Geometry

Vectors and Transfomations

Mensuration

Sets

Probability

Statistics

Calculator Skills