Constructing Triangles

Construct a triangle accurately with a ruler, compasses, a protractor and a set square: from its three sides, from two sides and the angle between them, or from two angles and a side. Then measure the parts you were not given.

Before you start

The three angles of a triangle add up to 180°, written (∠ sum of △). If two angles of a triangle are 55° and 80°, the third is $180^{\circ} - 55^{\circ} - 80^{\circ} = 45^{\circ}$ (the lesson Triangles and Their Angles).
In $\triangle ABC$ the side $BC$ is opposite $\angle BAC$ : it is the side that does not touch the vertex $A$ .
You have drawn lines of a given length with a ruler and angles of a given size with a protractor since Primary school. Measure a length from the 0 mark of the ruler, not from its end.

Instruments and the Words

In this lesson you draw triangles accurately from facts given about them: their lengths and angles, called the given data. You need a sharp pencil and these instruments.

Instrument	What it is used for
Ruler	Drawing straight lines; measuring lengths in centimetres, to the nearest 0.1 cm (1 mm).
Compasses	Drawing arcs (parts of circles). Every point on an arc is the same distance from its centre, so compasses mark off equal lengths.
Protractor	Measuring and drawing angles, to the nearest degree.
Set square	Drawing right angles (and parallel lines, in the lesson Constructing Quadrilaterals).

Construct and sketch

To construct a figure is to draw it accurately with instruments from the given data. Leave the construction arcs showing: they show how the figure was made.

A sketch is a quick drawing by hand, not to size, with the given data marked on it. Always sketch first: it shows where each length and each angle goes before you begin.

A protractor has two scales. To draw an angle of 50° at $P$ on the line $PQ$ , put the centre of the protractor on $P$ with its base line along $PQ$ . Use the scale that reads 0 on $PQ$ (here the inner one), count up to 50 and mark a dot; then join $P$ to the dot with a ruler. The other scale reads 130 at the same place.

Check the angle by eye: 50° is acute, smaller than a right angle. If the angle you drew is obtuse, you read the wrong scale.

Measuring and checking

Measure lengths to the nearest 0.1 cm and angles to the nearest degree.

Check yourself: before you measure the part you are asked for, measure one of the given lengths or angles on your drawing again. If it is not what was given, construct the triangle again more carefully.

Even a careful construction is a little out. A length you measure is right if it is within 0.2 cm of the answer, and an angle within 2°.

The drawings on this page

The drawings on this page are not to the real size: do not measure the screen. Construct each triangle on paper and measure your own drawing.

Three Sides

Draw one of the sides first; call it the base. The third vertex is a known distance from each end of the base. Every point on an arc of radius 6 cm about $A$ is 6 cm from $A$ , so the third vertex lies on that arc; it also lies on an arc about the other end. It is where the two arcs cross.

The arcs cross once above the base and once below it. Either crossing gives the same triangle turned over, so draw the arcs on one side only.

Example 1: Three sides

Construct $\triangle ABC$ in which $AB = 7$ cm, $BC = 5$ cm and $AC = 6$ cm. Measure $\angle ACB$ .

Solution

Sketch the triangle first and mark the three lengths on it.

Step 1. Draw the base $AB = 7$ cm with the ruler. Label its ends $A$ and $B$ .
Step 2. Set the compasses to 6 cm (the length $AC$ ) against the ruler. With the point on $A$ , draw an arc above $AB$ .
Step 3. Set the compasses to 5 cm (the length $BC$ ). With the point on $B$ , draw an arc that crosses the first arc. Label the crossing $C$ .
Step 4. Join $AC$ and $BC$ with the ruler. Leave the arcs showing.

Check yourself: measure $AC$ . It should be 6 cm.

Now measure $\angle ACB$ with the protractor, centre on $C$ , base line along $CA$ or $CB$ .

$\angle ACB \approx 78^{\circ}$ (any answer from 76° to 80° is right).

Two Sides and the Angle Between Them

When you know two sides and the angle between them (the included angle), draw one of the sides, draw the angle at the end of it where the two sides meet, and mark off the second side along the new arm.

The angle must be the one between the two given sides, at the vertex they share: the angle between $PQ$ and $PR$ is $\angle QPR$ , at $P$ .

Example 2: Two sides and the included angle

Construct $\triangle PQR$ in which $PQ = 6.5$ cm, $\angle QPR = 50^{\circ}$ and $PR = 5$ cm. Measure $QR$ .

Solution

Step 1. Draw $PQ = 6.5$ cm.
Step 2. Put the centre of the protractor on $P$ and its base line along $PQ$ . Mark 50° on the scale that reads 0 on $PQ$ , and draw a long line from $P$ through the mark.
Step 3. Set the compasses to 5 cm and, with the point on $P$ , draw a small arc across the arm (or measure 5 cm along it with the ruler). Label the point $R$ .
Step 4. Join $QR$ .

Check yourself: measure $\angle QPR$ . It should be 50°.

Measure $QR$ with the ruler.

$QR \approx 5.0$ cm (any answer from 4.8 cm to 5.2 cm is right).

Two Angles and the Side Between Them

Draw the given side. At each end of it draw the angle that belongs at that end, both on the same side of that line. The two new arms cross at the third vertex, so draw them long enough to cross.

Example 3: Two angles and the side between them

Construct $\triangle XYZ$ in which $XY = 7$ cm, $\angle YXZ = 40^{\circ}$ and $\angle XYZ = 65^{\circ}$ . Measure $XZ$ and $YZ$ .

Solution

The side $XY$ joins the vertices of the two given angles, $X$ and $Y$ : it is the side between them.

Step 1. Draw $XY = 7$ cm.
Step 2. With the protractor centred on $X$ and its base line along $XY$ , draw a long arm from $X$ at 40° to $XY$ .
Step 3. With the protractor centred on $Y$ and its base line along $YX$ , draw a long arm from $Y$ at 65° to $YX$ , on the same side of $XY$ as the first arm. Label the point where the two arms cross $Z$ .
Step 4. $XYZ$ is the triangle.

Check yourself: measure $\angle XYZ$ . It should be 65°, as given. The third angle, $\angle XZY$ , should be $180^{\circ} - 40^{\circ} - 65^{\circ} = 75^{\circ}$ (∠ sum of △).

Measure $XZ$ and $YZ$ with the ruler.

$XZ \approx 6.6$ cm (6.4 cm to 6.8 cm) and $YZ \approx 4.7$ cm (4.5 cm to 4.9 cm).

Two Angles and a Side Not Between Them

Sometimes the given side is not between the two given angles: one of the angles is at the vertex opposite the given side. You cannot draw that angle until its vertex is found.

So find the third angle first, with the angle sum of a triangle. Then you know the angles at both ends of the given side, and you construct as in the last section.

Example 4: Two angles and a side not between them

Construct $\triangle DEF$ in which $DE = 6$ cm, $\angle EDF = 48^{\circ}$ and $\angle DFE = 72^{\circ}$ . Measure $DF$ and $EF$ .

Solution

The side $DE$ joins $D$ and $E$ , but the second given angle is at $F$ . Find the angle at $E$ first:

$\angle DEF = 180^{\circ} - 48^{\circ} - 72^{\circ} = 60^{\circ}$ (∠ sum of △)

Now $DE$ is the side between two known angles, 48° at $D$ and 60° at $E$ .

Step 1. Draw $DE = 6$ cm.
Step 2. At $D$ , draw a long arm at 48° to $DE$ .
Step 3. At $E$ , draw a long arm at 60° to $ED$ , on the same side of $DE$ as the first arm. Label the crossing $F$ .
Step 4. Check yourself: measure $\angle DFE$ . It should be 72°, as given.

Measure $DF$ and $EF$ .

$DF \approx 5.5$ cm (5.3 cm to 5.7 cm) and $EF \approx 4.7$ cm (4.5 cm to 4.9 cm).

Right-Angled Triangles

In a right-angled triangle the side opposite the right angle is called the hypotenuse. It is the longest side.

Draw the right angle with the protractor (90°), or with a set square: lay one edge of the set square along the line, with its right-angled corner on the point, and draw along the other edge.

If you are given the two sides about the right angle, this is two sides and the angle between them: draw one side, draw the right angle at the end where the two sides meet, and mark off the other side.

Example 5: The two sides about the right angle

Construct $\triangle ABC$ in which $\angle ABC = 90^{\circ}$ , $AB = 4$ cm and $BC = 6.5$ cm. Measure $AC$ and $\angle ACB$ .

Solution

Step 1. Draw $BC = 6.5$ cm.
Step 2. At $B$ , draw a line at right angles to $BC$ , with the protractor or a set square.
Step 3. Set the compasses to 4 cm and, with the point on $B$ , mark the upright line. Label the point $A$ .
Step 4. Join $AC$ .

Check yourself: measure $\angle ABC$ . It should be 90°.

Measure $AC$ and $\angle ACB$ .

$AC \approx 7.6$ cm (7.4 cm to 7.8 cm) and $\angle ACB \approx 32^{\circ}$ (30° to 34°).

If you are given the hypotenuse and one other side, draw the other side first, and draw the right angle at the end of it where the triangle's right angle is ( $Q$ in Example 6). The third vertex is on that upright line, at a distance equal to the hypotenuse from the other end of the side you drew: an arc with that radius about that end finds it.

Example 6: The hypotenuse and one other side

Construct $\triangle PQR$ in which $\angle PQR = 90^{\circ}$ , $QR = 5$ cm and $PR = 8$ cm. Measure $PQ$ and $\angle PRQ$ .

Solution

$PR$ is opposite the right angle at $Q$ , so $PR$ is the hypotenuse.

Step 1. Draw $QR = 5$ cm.
Step 2. At $Q$ , draw a long line at right angles to $QR$ .
Step 3. Set the compasses to 8 cm. With the point on $R$ , draw an arc that cuts the upright line. Label the point $P$ .
Step 4. Join $PR$ .

Check yourself: measure $PR$ . It should be 8 cm.

Measure $PQ$ and $\angle PRQ$ .

$PQ \approx 6.2$ cm (6.0 cm to 6.4 cm) and $\angle PRQ \approx 51^{\circ}$ (49° to 53°).

Given	How to construct the triangle
Three sides	Draw one side; an arc about each end with the other two lengths; join to where the arcs cross.
Two sides and the angle between them	Draw one side; the angle at the end the two sides share; mark off the second side along the arm.
Two angles and the side between them	Draw the side; the angle that belongs at each end, both on the same side of it; the arms cross at the third vertex.
Two angles and a side not between them	Find the third angle (∠ sum of △); then as in the row above.
A right angle and the two sides about it	Draw one side; a right angle at the end where the two sides meet; mark off the other side.
A right angle, the hypotenuse and one other side	Draw the other side; a right angle at the vertex where it belongs; an arc about the other end with radius the hypotenuse.

When No Triangle Can Be Drawn

Try to construct a triangle with sides 3 cm, 4 cm and 8 cm. Draw the 8 cm side, an arc of radius 3 cm about one end and an arc of radius 4 cm about the other. The arcs do not meet: $3 + 4 = 7$ , which is less than 8, so the two short sides cannot reach each other.

With 3 cm, 5 cm and 8 cm the arcs meet only on the 8 cm line itself, since $3 + 5 = 8$ . The three "sides" lie flat along one line: there is no triangle.

Two angles can fail too. If the angles at the ends of a side add up to 180° or more, the arms do not cross on that side: with 100° and 85° they spread apart.

3 cm, 4 cm, 8 cm: the arcs do not meet.

3 cm, 5 cm, 8 cm: the arcs meet only on the line.

100° and 85°: the arms do not meet.

When a triangle can be drawn

From three lengths: the two shorter lengths must add up to more than the longest. If they add up to the longest or less, no triangle can be drawn.

From angles: two angles of a triangle must add up to less than 180°. If they add up to 180° or more, no triangle can be drawn.

Example 7: Can it be drawn?

Can a triangle be constructed from these data? Give a reason.

(a) Sides 4 cm, 5 cm and 10 cm. (b) Sides 6 cm, 7 cm and 9 cm. (c) $AB = 6$ cm, $\angle CAB = 95^{\circ}$ and $\angle CBA = 85^{\circ}$ .

Solution

(a) The two shorter sides: $4 + 5 = 9$ , which is less than the longest side, 10. The arcs would not meet.

(b) The two shorter sides: $6 + 7 = 13$ , which is more than 9. The arcs meet, so the triangle can be drawn.

(c) $95^{\circ} + 85^{\circ} = 180^{\circ}$ . The arms at $A$ and $B$ would never meet (they would be parallel), and the third angle would be $0^{\circ}$ .

(a) No, since $4 + 5 < 10$ . (b) Yes, since $6 + 7 > 9$ . (c) No, since the two angles add up to 180°.

Common Mistakes

Reading the wrong scale of the protractor, so that 50° is drawn as 130°. Use the scale that reads 0 on the line you start from, and check by eye: an acute angle must look acute.
Rubbing out the arcs. They are part of the construction and show how it was made: leave them.
Setting the compasses from the end of the ruler instead of its 0 mark, or letting them slip. Set the radius against the ruler and check it again before you draw each arc.
For two angles and a side not between them, drawing the two given angles at the two ends of the given side. One of them belongs at the vertex opposite the given side: find the third angle first.
Drawing the arms too short, so they end before they cross. Draw arms and arcs long enough to meet.
Thinking any three lengths make a triangle. With 3 cm, 4 cm and 8 cm the arcs never meet, because $3 + 4 < 8$ .
Measuring the drawings on the screen. They are not to the real size: construct on paper.

Practice

In each question, sketch the triangle first and mark the given data on it; then construct it on paper, leaving the arcs showing. Do the "check yourself" step before you measure. Give lengths to the nearest 0.1 cm and angles to the nearest degree: your answer is right if it is within 0.2 cm or 2° of the answer given.

Practice 1: Three sides

Sketch, then construct, $\triangle ABC$ in which $AB = 9$ cm, $BC = 6$ cm and $CA = 5$ cm.

Check yourself: measure $CA$ . Then measure $\angle ACB$ and say whether it is acute, right or obtuse.

ShowHide solution

Draw $AB = 9$ cm. With the compasses set to 5 cm, draw an arc about $A$ ; with them set to 6 cm, draw an arc about $B$ to cross it at $C$ . Join $AC$ and $BC$ .

Check: $CA$ measures 5 cm. Then measure $\angle ACB$ .

$\angle ACB \approx 109^{\circ}$ (107° to 111°): an obtuse angle.

Practice 2: Three sides, then a look at one angle

Sketch, then construct, $\triangle PQR$ in which $PQ = 6$ cm, $QR = 8$ cm and $PR = 10$ cm.

Check yourself: measure $PQ$ . Then measure $\angle PQR$ . What do you notice about $\angle PQR$ ?

ShowHide solution

Draw $QR = 8$ cm. Draw an arc of radius 6 cm about $Q$ and an arc of radius 10 cm about $R$ ; they cross at $P$ . Join $PQ$ and $PR$ .

Check: $PQ$ measures 6 cm. Then measure $\angle PQR$ .

$\angle PQR \approx 90^{\circ}$ (88° to 92°): it appears to be a right angle, so $\triangle PQR$ appears to be right-angled, with $PR$ the longest side. Measuring can only show that an angle is close to 90°, never that it is exactly 90°.

Practice 3: Two sides and an obtuse included angle

Sketch, then construct, $\triangle KLM$ in which $KL = 7.5$ cm, $\angle KLM = 110^{\circ}$ and $LM = 4.5$ cm.

Check yourself: measure $LM$ . Then measure $KM$ .

ShowHide solution

Draw $KL = 7.5$ cm. With the protractor centred on $L$ and its base line along $LK$ , mark 110° on the scale that reads 0 on $LK$ and draw a long arm from $L$ . The angle is obtuse, so the arm leans away from $K$ .

Mark $M$ on the arm with an arc of radius 4.5 cm about $L$ . Join $KM$ .

Check: $LM$ measures 4.5 cm. Then measure $KM$ .

$KM \approx 10.0$ cm (9.8 cm to 10.2 cm).

Practice 4: Is it equilateral?

Sketch, then construct, $\triangle ABC$ in which $BC = 7$ cm, $\angle ABC = 60^{\circ}$ and $AB = 7$ cm.

Check yourself: measure $AB$ . Then measure $AC$ , $\angle BAC$ and $\angle ACB$ . Is the triangle equilateral? Give a reason.

ShowHide solution

Draw $BC = 7$ cm. At $B$ draw a long arm at 60° to $BC$ . Mark $A$ on it with an arc of radius 7 cm about $B$ . Join $AC$ .

Check: $AB$ measures 7 cm. Then measure $AC$ and the two other angles.

This fits the lesson Isosceles and Equilateral Triangles: $AB = BC$ , so $\angle BAC = \angle ACB$ (base ∠s of isos. △), and each is $(180^{\circ} - 60^{\circ}) \div 2 = 60^{\circ}$ (∠ sum of △). All three angles are 60°, so the triangle is equilateral (equilateral △).

$AC \approx 7.0$ cm (6.8 cm to 7.2 cm), $\angle BAC \approx 60^{\circ}$ and $\angle ACB \approx 60^{\circ}$ (58° to 62°). Yes: all three angles are 60°, so the triangle is equilateral (equilateral △), and the measurements agree.

Practice 5: Two angles and the side between them

Sketch, then construct, $\triangle PQR$ in which $QR = 8$ cm, $\angle PQR = 35^{\circ}$ and $\angle PRQ = 75^{\circ}$ .

Check yourself: measure $\angle PRQ$ . Then measure $PQ$ and $PR$ .

ShowHide solution

Draw $QR = 8$ cm. At $Q$ draw a long arm at 35° to $QR$ ; at $R$ draw a long arm at 75° to $RQ$ . They cross at $P$ .

Check: $\angle PRQ$ measures 75°, and the third angle $\angle QPR$ should be $180^{\circ} - 35^{\circ} - 75^{\circ} = 70^{\circ}$ (∠ sum of △). Then measure $PQ$ and $PR$ .

$PQ \approx 8.2$ cm (8.0 cm to 8.4 cm) and $PR \approx 4.9$ cm (4.7 cm to 5.1 cm).

Practice 6: Two angles and a side not between them

Sketch, then construct, $\triangle STU$ in which $ST = 6$ cm, $\angle TSU = 62^{\circ}$ and $\angle SUT = 50^{\circ}$ .

Check yourself: measure $\angle SUT$ . Then measure $SU$ and $TU$ .

ShowHide solution

The side $ST$ is not between the angles at $S$ and $U$ . Find the angle at $T$ first:

$\angle STU = 180^{\circ} - 62^{\circ} - 50^{\circ} = 68^{\circ}$ (∠ sum of △)

Draw $ST = 6$ cm. At $S$ draw a long arm at 62° to $ST$ ; at $T$ draw a long arm at 68° to $TS$ . They cross at $U$ .

Check: $\angle SUT$ measures 50°, as given. Then measure $SU$ and $TU$ .

$SU \approx 7.3$ cm (7.1 cm to 7.5 cm) and $TU \approx 6.9$ cm (6.7 cm to 7.1 cm).

Practice 7: The two sides about the right angle

Sketch, then construct, $\triangle DEF$ in which $\angle DEF = 90^{\circ}$ , $DE = 5.5$ cm and $EF = 4$ cm.

Check yourself: measure $\angle DEF$ . Then measure $DF$ and $\angle EDF$ .

ShowHide solution

Draw $ED = 5.5$ cm. At $E$ draw a line at right angles to $ED$ (protractor or set square). Mark $F$ on it with an arc of radius 4 cm about $E$ . Join $DF$ .

Check: $\angle DEF$ measures 90°. Then measure $DF$ and $\angle EDF$ .

$DF \approx 6.8$ cm (6.6 cm to 7.0 cm) and $\angle EDF \approx 36^{\circ}$ (34° to 38°).

Practice 8: The hypotenuse and one other side

Sketch, then construct, $\triangle XYZ$ in which $\angle XYZ = 90^{\circ}$ , $XY = 6$ cm and $XZ = 9$ cm.

Check yourself: measure $XZ$ . Then measure $YZ$ and $\angle YXZ$ .

ShowHide solution

$XZ$ is opposite the right angle at $Y$ : it is the hypotenuse. Draw the other given side first.

Draw $YX = 6$ cm. At $Y$ draw a long line at right angles to $YX$ . With the compasses set to 9 cm and the point on $X$ , draw an arc cutting that line at $Z$ . Join $XZ$ .

Check: $XZ$ measures 9 cm. Then measure $YZ$ and $\angle YXZ$ .

$YZ \approx 6.7$ cm (6.5 cm to 6.9 cm) and $\angle YXZ \approx 48^{\circ}$ (46° to 50°).

Practice 9: Can it be drawn?

Without constructing, say whether a triangle can be drawn from each set of data, and give a reason.

(a) Sides 3 cm, 6 cm and 10 cm. (b) Sides 5 cm, 7 cm and 11 cm. (c) Sides 6 cm, 6 cm and 12 cm.

(d) $AB = 5$ cm, $\angle CAB = 110^{\circ}$ , $\angle CBA = 75^{\circ}$ . (e) $AB = 5$ cm, $\angle CAB = 100^{\circ}$ , $\angle CBA = 70^{\circ}$ .

ShowHide solution

(a) $3 + 6 = 9$ , less than 10: the two shorter sides cannot reach. No.

(b) $5 + 7 = 12$ , more than 11. Yes.

(d) $110^{\circ} + 75^{\circ} = 185^{\circ}$ , more than 180°: the arms spread apart. No.

(e) $100^{\circ} + 70^{\circ} = 170^{\circ}$ , less than 180°. Yes: the third angle is $180^{\circ} - 170^{\circ} = 10^{\circ}$ .

(a) No (b) Yes (c) No (d) No (e) Yes.

Practice 10: A flag for bunting

Priya is making a paper flag for a string of bunting for her class party. The flag is a triangle $ABC$ hanging from its top edge $AB = 10$ cm, with $\angle CAB = \angle CBA = 70^{\circ}$ and the point $C$ at the bottom.

Sketch, then construct, the flag full size. Check yourself: measure $\angle CBA$ . Then measure the two slanting edges $AC$ and $BC$ .

ShowHide solution

Draw $AB = 10$ cm near the top of the page. At $A$ and at $B$ draw long arms at 70° to $AB$ , going down the page. They cross at $C$ .

Check: $\angle CBA$ measures 70°. Then measure $AC$ and $BC$ .

The two angles at $A$ and $B$ are equal, so $AC = BC$ (isos. △): the two measurements should agree. The angle at the bottom is $180^{\circ} - 70^{\circ} - 70^{\circ} = 40^{\circ}$ .

$AC \approx BC \approx 14.6$ cm (14.4 cm to 14.8 cm).

Practice 11: A set square

Mr Tan wants a full-size drawing of a set square: a triangle $ABC$ with $\angle ACB = 90^{\circ}$ , $\angle BAC = 30^{\circ}$ and its longest side $AB = 10$ cm.

Sketch, then construct, the triangle. Check yourself: measure $\angle ACB$ . Then measure $BC$ and $AC$ .

ShowHide solution

The side $AB$ is not between the angles at $A$ and $C$ . Find the angle at $B$ first:

$\angle ABC = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$ (∠ sum of △)

Draw $AB = 10$ cm. At $A$ draw a long arm at 30° to $AB$ ; at $B$ draw a long arm at 60° to $BA$ . They cross at $C$ .

Check: $\angle ACB$ measures 90°. Then measure $BC$ and $AC$ .

$BC \approx 5.0$ cm (4.8 cm to 5.2 cm) and $AC \approx 8.7$ cm (8.5 cm to 8.9 cm).

Practice 12: A model ladder

For a science project, Ali leans a model ladder 10 cm long against an upright wall. The foot of the ladder $G$ is on flat ground, 4 cm from the foot of the wall $W$ , and the top of the ladder $T$ touches the wall.

Sketch the triangle $WGT$ and mark the right angle. Construct it. Check yourself: measure $GT$ . Then measure how high up the wall the ladder reaches ( $WT$ ) and the angle the ladder makes with the ground ( $\angle TGW$ ).

ShowHide solution

The wall is upright on flat ground, so $\angle GWT = 90^{\circ}$ , and the ladder $GT = 10$ cm is the hypotenuse.

Draw $WG = 4$ cm. At $W$ draw a long line at right angles to $WG$ (the wall). With the compasses set to 10 cm and the point on $G$ , draw an arc cutting the wall at $T$ . Join $GT$ .

Check: $GT$ measures 10 cm. Then measure $WT$ and $\angle TGW$ .

The ladder reaches $WT \approx 9.2$ cm up the wall (9.0 cm to 9.4 cm) and makes $\angle TGW \approx 66^{\circ}$ (64° to 68°) with the ground.

Key Points

To construct is to draw accurately with instruments from the given data; leave the arcs showing. Sketch first, with the data marked.
Three sides: draw one side, then an arc about each end; the third vertex is where the arcs cross.
Two sides and the included angle: draw one side, the angle at the shared vertex, then mark off the other side.
Two angles and the side between them: draw the side and the angle at each end; the arms cross at the third vertex.
Two angles and a side not between them: find the third angle first (∠ sum of △).
Right-angled: draw the right angle with a protractor or set square; given the hypotenuse, find the third vertex with an arc of that radius about the other end of the side you drew.
No triangle can be drawn if the two shorter sides add up to the longest or less, or if two angles add up to 180° or more.
Check yourself by measuring a given length or angle again; measure lengths to 0.1 cm and angles to the nearest degree, within 0.2 cm or 2°.