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Drawing triangles from given sides and angles: one, many or none

You are told some of a triangle’s sides and angles. Can you draw it, and is it the only triangle that fits?

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Triangle inequalityMath 7
Where you are 18 in Triangles
Geometry11
Triangles18
▴ 4 aboveCongruent triangles3Triangle inequality
Drawing triangles from given sides and angles: one, many or none
SSS, SAS, ASA, AASSides and angles in order▾ Show all 18 (9 more)

When you reach for it

A plan for a triangle never lists everything about it. A garden plan gives the three edges of a bed; a tent plan gives two edges of a panel and the angle between them; a roof truss is given by its span and the slope of the roof at each end. Before anything is cut, two questions need answers: can a triangle with these measurements exist, and is it the only one? If two different triangles fit, the plan has not said enough.

The way to find out is to draw it with a ruler, a protractor and a compass, one piece at a time, watching at each step whether you had a choice. Every set of conditions ends in one of three ways.

  • Exactly one triangle. Every triangle that fits can be cut out and laid exactly on any other, perhaps after turning it over. Where it sits on the page and which way it faces do not count. Triangles that match like this are congruent.
  • Many triangles. Triangles that really differ, in shape or in size, all fit.
  • No triangle. The pieces cannot be made to close.

Part 1, the first day, takes the conditions that always settle the question: three sides, two sides with the angle between them, and two angles with the side between them. Part 2, the second day, takes the rest: three angles, two angles with a side not between them, and two sides with an angle not between them.

The procedure

The three tools. A ruler draws and measures a length, a protractor an angle. A compass set to 5 cm with its point on A draws points that are all exactly 5 cm from A. All such points lie on one circle, and a short piece of it is an arc. So “the corner is 5 cm from A” means “the corner is on this circle”, and the arc shows every place the corner could be.

Between. In \triangle ABC the side AB is between the angles at A and B, because it joins their vertices. The angle at A is between the sides AB and AC, because both start at A. The side BC is opposite the angle at A. Which parts are between which (the parts of a triangle) decides how the drawing goes.

Taken as known. The three angles of a triangle add to 180^\circ (the triangle angle sum), and any two sides together are longer than the third (the triangle inequality).

Part 1: conditions that always settle it

Three sides. Draw the longest side and call its ends A and B. The third corner C is the second length away from A, so it lies on the circle of that radius about A: draw an arc of it. It is the third length away from B, so it also lies on a circle about B: draw a second arc. C goes where the arcs cross. Join it to A and to B.

Two circles cross in at most two points, and here the two crossings sit one on each side of AB, mirror images of each other. Fold the page along AB and one triangle lands on the other. So three sides give exactly one triangle, whenever they give one at all.

They give none when the arcs fail to cross. The arcs reach toward each other along AB from its two ends, and they cross only when the two shorter sides together are longer than the longest. If together they are shorter, a gap is left. If together they are exactly as long, the arcs meet on AB itself and the three corners lie in a straight line, a flat figure that is not a triangle. That test is the triangle inequality, and with the longest side as the base it is the only sum to check.

Two sides and the angle between them. Draw one side, AB. Put the protractor’s center on A with its baseline along AB, mark the angle, and draw a ray from A through the mark. The other side also starts at A, since the angle is between the two sides, so measure it along this ray and mark C. Join B to C.

Every step was forced: BC just joins two points already fixed, and the only free choice, which side of AB to draw the ray on, makes a mirror image. So two sides and the angle between them give exactly one triangle, and always one: nothing has to meet, so any two lengths and any angle between 0^\circ and 180^\circ will do.

Two angles and the side between them. Draw the side AB, then the angle at A and the angle at B, both opening on the same side of AB. The corner C is where the two new rays cross, and two rays cross at most once, so two angles and the side between them give exactly one triangle, as long as the rays do cross.

They cross exactly when the two angles add to less than 180^\circ, because the angle at C would be 180^\circ minus the two given angles, and an angle of a triangle is more than 0^\circ. If they add to 180^\circ the rays run side by side and never meet; if they add to more, the rays lean apart. Either way there is no triangle.

Part 2: conditions that may not settle it

Three angles. Add them first. If the sum is not 180^\circ, no triangle has these angles. If it is, choose any length for one side and draw two of the angles at its ends, as in Part 1; the third comes out right by the angle sum. But the length was your choice. A longer side gives a larger triangle of the same shape, a scale copy of the first. So three angles that add to 180^\circ give many triangles, one shape in every size. Angles settle the shape; only a length settles the size.

Angle sets that no triangle has. Two obtuse angles already add to more than 180^\circ. Two right angles make 180^\circ and leave 0^\circ for the third. A right angle and an obtuse one add to more than 180^\circ. So a triangle has at most one angle that is right or obtuse, and any set with two such angles gives no triangle.

Two angles and a side not between them. Find the third angle from the angle sum. The given side joins two corners whose angles are now both known, so it is between two known angles and the drawing goes as in Part 1: exactly one triangle, provided the two given angles add to less than 180^\circ.

Two sides and an angle not between them. Say the angle is at A, the side AB lies along one arm, and the other given side is BC, opposite A. Draw the angle as two rays and measure AB along one of them. The corner C must be on the other ray and BC away from B, so set the compass to BC, put its point on B, and swing an arc across the other ray. Each crossing is a place C can go.

When the angle at A is acute, the count depends on one more measurement: the shortest distance from B to the other ray, straight down to it at a right angle.

  • If BC is shorter than that distance, the arc stops short of the ray: no triangle.
  • If BC equals it, the arc just touches the ray, making a right angle at C: one triangle.
  • If BC is longer than that distance but shorter than AB, the arc crosses the ray twice: two different triangles. They are not mirror images: both have the given angle, the given AB and the given BC, yet their third sides differ.
  • If BC is as long as AB or longer, one crossing is on the ray and the other lands at A or behind it, off the ray: one triangle.

So this arrangement can give two, one or none, which is why it is not one of the conditions that always fix a triangle (side, side, angle). A right or obtuse angle at A is simpler, as Boundaries explains.

Worked examples

Examples 1 to 4 go with Part 1, and Examples 5 to 7 with Part 2.

Example 1

A garden bed from three edges

Theodosia is planning a triangular garden bed with edges of 7 m, 5 m and 4 m, drawn at 1 cm for every meter. Construct her drawing, decide how many different beds have these edges, and measure the largest angle.

Solution

The drawing has sides 7 cm, 5 cm and 4 cm. The two shorter sides make 5 + 4 = 9 cm, more than 7 cm, so the arcs will cross.

Draw AB = 7 cm. With the compass at 5 cm and its point on A, draw an arc above AB and one below. With it at 4 cm and its point on B, draw arcs that cross the first two. Call the upper crossing C and join it to A and B.

5 cm 4 cm 7 cm A B C C′
Figure 1The third vertex is 5 cm from A and 4 cm from B, so it sits where the two arcs cross. The crossing below, C', gives the same triangle turned over.

The lower crossing C' is the mirror image of C in AB, so it gives the same triangle turned over: there is exactly one garden bed. The largest angle is opposite the longest side, at C. The protractor reads about 102^\circ, an obtuse corner.

Example 2

When the arcs miss

Which of these sets of lengths are the sides of a triangle?

(a) 9 cm, 4 cm and 4 cm \qquad (b) 9 cm, 5 cm and 4 cm \qquad (c) 9 cm, 6 cm and 4 cm

Solution

In each case draw AB = 9 cm, an arc of 4 cm about B, and the other arc about A.

(a) The arc about A meets AB at 4 cm from A, and the one about B at 4 cm from B. Together they cover 4 + 4 = 8 cm of the 9 cm side, leaving a gap of 1 cm. The arcs never meet: no triangle.

(b) The arcs reach 5 cm and 4 cm along AB, and 5 + 4 = 9: they meet on the segment itself, 5 cm from A. C would be on AB, so the figure is flat: no triangle.

(c) 6 + 4 = 10 is more than 9, so the arcs overlap and cross above and below AB: exactly one triangle.

Example 3

A tent panel from two edges and the angle between them

Umberto is cutting a triangular tent panel. On his plan two edges measure 7 cm and 5 cm, with an angle of 40^\circ between them. Construct the panel, say how many panels fit the plan, and measure the third edge and the angle opposite the 7 cm edge.

Solution

Draw AB = 7 cm. With the protractor’s center on A and its baseline along AB, mark 40^\circ and draw the ray from A through the mark. The 5 cm edge also starts at A, because the angle is between the two edges, so measure 5 cm along the ray and mark C. Join B to C.

40° 5 cm 7 cm A B C
Figure 2The angle at A and the two sides beside it leave nothing to choose. BC is whatever joins the two points already fixed.

Every step was forced, so there is exactly one panel: anyone following the plan cuts the same one. Measuring, BC is about 4.5 cm and the angle at C is about 95^\circ.

Example 4

A roof truss from its span and two end angles

Wendell draws a roof truss at 1 cm for every meter. (a) The span is 8 m, and the roof meets it at 30^\circ at one end and 45^\circ at the other. Construct the truss, say how many fit, and find the angle at the top and the lengths of the two sloping pieces. (b) He then tries angles of 100^\circ and 85^\circ at the ends of the same span. What happens?

Solution

(a) Draw AB = 8 cm. At A draw a ray at 30^\circ to AB, and at B a ray at 45^\circ to BA, both above the line. They cross at C.

30° 45° 105° 8 cm A B C
Figure 3Two rays cross at most once, so the span and its two end angles fix the truss.

Two rays cross at most once, so there is exactly one truss. The angle at the top needs no measuring, by the angle sum:

m\angle C = 180^\circ - 30^\circ - 45^\circ = 105^\circ

and the protractor agrees. Measuring, AC is about 5.9 cm and BC is about 4.1 cm, so the sloping pieces are about 5.9 m and 4.1 m long.

(b) 100^\circ + 85^\circ = 185^\circ, more than 180^\circ. The rays lean away from each other and never meet: no triangle, so no truss.

Example 5

Same angles, different sizes

Valentina and Xenia each draw a triangle with angles of 50^\circ, 55^\circ and 75^\circ. Valentina makes the side between the 50^\circ and 55^\circ angles 4 cm long, and Xenia makes it 6 cm long. Are their triangles the same?

Solution

50^\circ + 55^\circ + 75^\circ = 180^\circ, so these can be the angles of a triangle. Each draws her side with the 50^\circ and 55^\circ angles at its ends, as in Part 1, and the third corner comes out as 75^\circ without being drawn on purpose.

50° 55° 75° 4 cm 50° 55° 75° 6 cm
Figure 4The same three angles on bases of 4 cm and 6 cm. Same shape, different sizes.

Measuring, Valentina’s other two sides are about 3.2 cm and 3.4 cm, and Xenia’s are about 4.8 cm and 5.1 cm. The triangles are not the same: Xenia’s is larger. They do have the same shape. Xenia’s base is 6 \div 4 = 1.5 times Valentina’s, and so is each of her other sides: 3.2 \times 1.5 = 4.8 and 3.4 \times 1.5 = 5.1. Any other length for that side gives yet another triangle with these angles, so the angles give many triangles.

Example 6

Angle sets that work, and ones that cannot

Say whether each description gives one triangle, many, or none.

(a) Angles of 48^\circ, 62^\circ and 70^\circ.

(b) A triangle with two angles of 95^\circ and 100^\circ.

(c) Angles of 90^\circ, 57^\circ and 43^\circ.

(d) Angles of 64^\circ and 76^\circ, with the side opposite the 64^\circ angle 5 cm long.

Solution

(a) 48^\circ + 62^\circ + 70^\circ = 180^\circ. Three angles with the right sum give many triangles, one shape in every size.

(b) Two obtuse angles: 95^\circ + 100^\circ = 195^\circ is already more than 180^\circ, before the third angle is counted. None.

(c) 90^\circ + 57^\circ + 43^\circ = 190^\circ, not 180^\circ. None.

(d) The third angle is 180^\circ - 64^\circ - 76^\circ = 40^\circ. The side opposite the 64^\circ angle joins the other two corners, the ones with 76^\circ and 40^\circ, so it is between two known angles. Draw the 5 cm side with 76^\circ at one end and 40^\circ at the other, as in Part 1: exactly one triangle.

Example 7

Two sides and an angle not between them

In \triangle ABC, m\angle A = 32^\circ, AB = 6 cm, and BC is the side opposite \angle A. How many triangles are there when (a) BC = 4 cm, (b) BC = 2.5 cm, (c) BC = 7 cm? Where there is a triangle, measure AC.

Solution

Draw a 32^\circ angle at A with one ray along the bottom, and mark B on the other ray, 6 cm from A. The corner C has to be on the bottom ray, at the given distance from B. The angle is acute, so first measure the shortest distance from B to the bottom ray, straight down at a right angle: it is about 3.2 cm.

(a) 4 cm is more than 3.2 cm and less than AB = 6 cm, so the arc of 4 cm about B crosses the bottom ray twice, at C_1 and C_2.

32° 6 cm B 4 cm 4 cm A
C_1
C_2
Figure 5An arc of 4 cm about B crosses the ray twice. Triangles ABC_1 and ABC_2 both fit the conditions.

AC_1 is about 2.7 cm and AC_2 about 7.5 cm. Both triangles have the 32^\circ angle, the 6 cm side and the 4 cm side, but their third sides differ, so there are two triangles. The angle at C_1 is obtuse and the one at C_2 acute, another sign they cannot be laid on each other.

(b) 2.5 cm is less than the 3.2 cm the ray keeps away from B, so the arc stops short: no triangle.

32° 6 cm B 3.2 cm A
Figure 6The ray never comes closer to B than about 3.2 cm, so an arc of 2.5 cm cannot reach it.

(c) 7 cm is longer than AB = 6 cm. The arc crosses the bottom ray once, at C, with AC about 11.3 cm. Its other crossing, D, is on the line behind A, off the ray, so it gives no triangle with a 32^\circ angle at A: one triangle.

32° 6 cm B 7 cm A C D
Figure 7An arc of 7 cm crosses the ray once, at C. Its other crossing, D, is behind A and not on the ray.

Same angle, same AB, three different answers. Only the length of BC changed.

Boundaries

Lengths must be positive and angles strictly between 0^\circ and 180^\circ. A side of 0 cm is a point and an angle of 180^\circ is a straight line; neither can be part of a triangle.

A drawing is only as exact as the tools. A ruler reads to about a millimeter and a protractor to about a degree, so every measured answer here is “about”. Right at an edge a drawing cannot decide: sides of 9 cm, 5 cm and 4.05 cm do make a triangle, but one so flat that a pencil line hides it. The exact tests, the triangle inequality and the angle sum, decide those cases.

At least one length is needed. However many angles are given, they fix only the shape, so a triangle with no length given comes in every size.

Two sides and a right or obtuse angle not between them. Then the point of the other ray closest to B is A itself, so the arc can cross the ray only beyond A, and only once: one triangle when BC is longer than AB, none when it is not.

Flat paper. Everything here, the angle sum included, is about triangles drawn on a flat surface.

Where this goes next

The conditions that gave exactly one triangle (three sides, two sides and the angle between them, two angles and any one side) become the congruence criteria SSS, SAS, ASA and AAS. There they are proved rather than seen in a drawing, and used to show two triangles must match without measuring them. Why side, side, angle and three angles fall short is taken up in why SSA and AAA fail.

That three angles fix the shape but not the size is the start of similar triangles: two pairs of matching angles make two triangles scale copies of each other. And the compass moves used here are made exact in copying a segment and an angle.

Exercises

Part 1: three sides

For each set of sides, try to construct the triangle, and say whether the sides give one triangle or none.

  1. 8 cm, 6 cm and 5 cm. If there is a triangle, measure its largest angle.
  2. 2 cm, 3 cm and 6 cm.
  3. 4 cm, 7 cm and 11 cm.
  4. 6 cm, 6 cm and 6 cm. If there is a triangle, measure its angles.
  5. Ysolde wants a triangular kite frame made of three sticks 40 cm, 25 cm and 70 cm long. Can the frame be made? If not, how much shorter must the longest stick be cut?

Part 1: two sides and the angle between them

  1. Construct a triangle with sides of 6 cm and 4 cm and an angle of 70^\circ between them. How many triangles fit, and how long is the third side?
  2. Xenia says a triangle with sides of 3 cm and 9 cm and an angle of 120^\circ between them cannot be drawn, because 3 cm is too short. Is she right? Construct it if you can, and measure the third side.

Part 1: two angles and the side between them

Say whether each gives one triangle or none. Where there is one, find the third angle.

  1. A side of 7 cm with angles of 50^\circ and 60^\circ at its two ends.
  2. A side of 5 cm with angles of 110^\circ and 75^\circ at its two ends.
  3. A side of 6 cm with a right angle at each end.

Part 2: three angles

Say whether each set of angles gives many triangles or none.

  1. 25^\circ, 70^\circ and 85^\circ.
  2. 44^\circ, 66^\circ and 80^\circ.
  3. 90^\circ, 45^\circ and 45^\circ.
  4. Valentina and Umberto each draw a triangle with angles of 30^\circ, 70^\circ and 80^\circ. Valentina makes the side between the 30^\circ and 70^\circ angles 5 cm long, and Umberto makes it 8 cm long. Are the triangles the same? How many times as long as each of Valentina’s sides is the matching side of Umberto’s?

Part 2: two sides and an angle not between them

In Exercises 15 to 18, m\angle A = 30^\circ, AB = 8 cm, and BC is the side opposite \angle A. Draw the angle, mark B, and swing an arc of radius BC about B. Say how many triangles there are, and measure AC in each one.

  1. BC = 3 cm
  2. BC = 5 cm
  3. BC = 4 cm
  4. BC = 9 cm

Part 2: mixed conditions

  1. In \triangle ABC, m\angle A = 40^\circ, m\angle B = 75^\circ and BC = 6 cm. How many triangles fit? Find m\angle C.
  2. Theodosia designs a roof truss with a span of 9 m and the roof meeting the span at 35^\circ at both ends. She draws it at 1 cm for every meter. How many trusses fit, what is the angle at the top, and how long is each sloping piece?

Reasoning

  1. The two arcs in a three-sides construction usually cross twice. Explain why three sides still give only one triangle.
  2. Explain why no list of three angles, however carefully chosen, can fix a single triangle.
  3. Explain why no triangle has two obtuse angles, two right angles, or a right angle and an obtuse angle.
  4. In the two-sides-and-an-angle-not-between construction, with an acute angle at A, explain why there is exactly one triangle when BC is longer than AB.
Answers
  1. One triangle. 6 + 5 = 11 is more than 8, so the arcs cross. The largest angle, opposite the 8 cm side, measures about 93^\circ.
  2. No triangle. 2 + 3 = 5 is less than 6, so the arcs about the ends of the 6 cm side leave a gap of 1 cm.
  3. No triangle. 4 + 7 = 11 exactly, so the arcs meet on the 11 cm side itself and the figure is flat.
  4. One triangle. 6 + 6 = 12 is more than 6. Every angle measures 60^\circ.
  5. No: 40 + 25 = 65, which is less than 70, so the frame cannot close. The longest stick must be cut to less than 65 cm, so by more than 5 cm. (Cut it below 15 cm, though, and it is then too short instead, since 40 - 25 = 15.)
  6. Exactly one triangle, since every step is forced. The third side is about 6.0 cm.
  7. She is wrong. Two sides and the angle between them always give exactly one triangle, because nothing has to meet: the third side is just the line joining the two ends. It is about 10.8 cm.
  8. One triangle. 50^\circ + 60^\circ = 110^\circ is less than 180^\circ, and the third angle is 180^\circ - 110^\circ = 70^\circ.
  9. No triangle. 110^\circ + 75^\circ = 185^\circ, more than 180^\circ, so the rays lean apart.
  10. No triangle. 90^\circ + 90^\circ = 180^\circ, so both rays stand straight up from the side and run side by side without meeting.
  11. Many. 25^\circ + 70^\circ + 85^\circ = 180^\circ.
  12. None. 44^\circ + 66^\circ + 80^\circ = 190^\circ.
  13. Many. 90^\circ + 45^\circ + 45^\circ = 180^\circ: right triangles with two equal sides, in every size.
  14. They have the same shape but are not the same triangle, because Umberto’s is larger. Each of Umberto’s sides is 8 \div 5 = 1.6 times as long as the matching side of Valentina’s.
  15. No triangle. The shortest distance from B to the other ray is 4 cm, and an arc of 3 cm cannot reach it.
  16. Two triangles. 5 cm is more than 4 cm and less than 8 cm, so the arc crosses the ray twice: AC is about 3.9 cm in one and about 9.9 cm in the other.
  17. One triangle. The arc just touches the ray, where the angle at C is a right angle, and AC is about 6.9 cm.
  18. One triangle. 9 cm is longer than AB = 8 cm, so the second crossing is behind A, off the ray. AC is about 15.0 cm.
  19. One triangle. m\angle C = 180^\circ - 40^\circ - 75^\circ = 65^\circ. Now BC is between the 75^\circ angle at B and the 65^\circ angle at C, so the drawing is the side with two angles at its ends.
  20. Exactly one, since the span is between the two given angles. The angle at the top is 180^\circ - 35^\circ - 35^\circ = 110^\circ. Each sloping piece measures about 5.5 cm on the drawing, so it is about 5.5 m long.
  21. The arcs cross at most twice, once on each side of the base. The two crossings are mirror images of each other in the base, so the two triangles fit exactly on each other once one is turned over: they are the same triangle. Any other drawing with these sides can be moved so that its longest side lies on the same base, and then its third corner is one of the same two crossings.
  22. Draw any one side, at any length, and put two of the angles at its ends; the third comes out right by the angle sum. Nothing in the angles says how long that first side is, so a longer or shorter side gives a larger or smaller triangle with exactly the same angles. Only a length can fix the size.
  23. The three angles add to 180^\circ and each must be more than 0^\circ. Two obtuse angles are each more than 90^\circ, so they already add to more than 180^\circ. Two right angles add to 180^\circ, leaving 0^\circ for the third. A right angle and an obtuse angle add to more than 180^\circ. In each case there is nothing left for the third angle.
  24. A is only AB away from B, which is less than BC, so A lies inside the circle of radius BC about B. Walk along the line of the ray from A in either direction: you start closer to B than BC and end up as far from B as you like, so somewhere on each side of A you are exactly BC from B, on the circle. A line crosses a circle at most twice, so there is exactly one crossing on each side. The one beyond A is on the ray and gives a triangle; the one behind A is off the ray and gives none. That leaves exactly one triangle.
kg1128 · updated 2026-10-07 · CC-BY-SA-4.0