High School Maths · Year 2 · Unit 15 · Lesson 2 of 5

Constructing Triangles and Angles: Make 60°, 90°, 45° and 30° Without a Protractor

About 15 minutes · constructing triangles from SSS, SAS and ASA · the triangle inequality · constructing 60°, 90°, 45° and 30° · combining constructed angles · when two sides and an angle give two triangles · no calculator · exit quiz

Hook

You have a ruler and compasses, but no protractor.

Can you draw an angle of exactly 60°?

Yes. Draw a line and mark a point A. Draw an arc from A that cuts the line at B. Keep the same radius and draw an arc from B. Join A to the point C where the arcs cross. AB, BC and CA are all one radius long, so triangle ABC is equilateral and angle CAB is exactly 60°. Bisect it and you have 30°.

Learn

Master the essentials

SSS: draw one side, then an arc from each end. Where the arcs cross is the third corner.
Construct 90° (the perpendicular at P), then bisect it to get 45°.

SSS: three sides

Draw one side. From each end, draw an arc with the length of the side that starts there. Join to where the arcs cross.

AB = 7 cm, then arcs of 6.5 cm from A and 5 cm from B

SAS: two sides and the angle between

Draw one side. Measure the angle at one end with a protractor. Mark the second side along that arm and join up.

The angle must be BETWEEN the two given sides

ASA: two angles and the side between

Draw the side. Draw the given angle at each end. The two arms cross at the third corner.

The third angle is 180° minus the other two

Triangle inequality

The two shorter sides must add up to MORE than the longest side, or the arcs never meet.

3 + 4 > 6 ✓, but 3 + 4 < 8 ✗

60°

Arc from A cuts the line at B. Same radius: arc from B. Join A to the crossing point. The triangle is equilateral.

No protractor needed

90°

Construct the perpendicular at a point P on a line: arcs from P cut the line on each side, then bisect the gap between those points.

This is the perpendicular bisector again

45° and 30°

Bisect 90° to get 45°. Bisect 60° to get 30°. Add or subtract constructed angles for others.

Bisect 30° to get 15°; 90° + 30° = 120°

One triangle or two?

SSS, SAS and ASA each fix one triangle. Two sides and an angle that is NOT between them can give two different triangles.

Check where the angle is before you construct

Worked examples

See it done

1. Construct triangle ABC with AB = 8 cm, AC = 5 cm and BC = 6 cm. List the steps.

  1. Draw AB = 8 cm with a ruler.
  2. Open the compasses to 5 cm (the length of AC). Put the point on A and draw an arc above AB.
  3. Open them to 6 cm (the length of BC). Put the point on B and draw an arc that crosses the first arc. The crossing point is C.
  4. Join AC and BC. Leave the arcs on show.

Arc of radius 5 cm from A and arc of radius 6 cm from B cross at C

2. Construct an angle of 45° at a point P on a straight line.

  1. From P, draw an arc that cuts the line on both sides of P.
  2. From these two points, draw equal, larger arcs that cross above P. Join P to the crossing point: this line is at 90° to the first line.
  3. Bisect the right angle: an arc from P cuts both arms, then equal arcs from those two points cross. Join P to that crossing point.
  4. The new line makes 90° ÷ 2 = 45° with the first line.

Construct 90°, then bisect it

3. Show that a triangle with sides of 4 cm, 7 cm and 12 cm cannot be constructed.

  1. Draw the longest side, 12 cm.
  2. The third corner must be 4 cm from one end and 7 cm from the other, so arcs of these radii must cross.
  3. 4 + 7 = 11, which is less than 12. Along the base, the arcs reach only 4 cm and 7 cm, leaving a gap of 1 cm, so they never meet.
  4. The two shorter sides must add up to more than the longest side.

4 + 7 = 11 < 12, so there is no triangle

Common mistake: Drawing each arc from the wrong end. In SSS, the arc from A has the length of a side that ENDS at A (AC), not BC. Write the length next to each arc as you draw it, and check the finished sides with a ruler (within 2 mm).

PractiseNo calculator

10 questions with instant feedback

Choose the correct answer. Sketch the construction first: which side or angle do you draw first, and where do the arcs go? Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.Ali draws a line through a point A. He draws an arc from A that cuts the line at B. With the same radius, he draws an arc from B that crosses the first arc at C. He joins A to C. What is the size of angle CAB?

Not quite. This is the angle made by the perpendicular construction.

Solution

  1. AB, BC and AC are all one radius long.
  2. So triangle ABC is equilateral.
  3. Every angle in an equilateral triangle is 60°.

Answer: D. 60°.

Not quite. This is half of a right angle.

Solution

  1. AB, BC and AC are all one radius long.
  2. So triangle ABC is equilateral.
  3. Every angle in an equilateral triangle is 60°.

Answer: D. 60°.

Not quite. This is half of the angle. Ali has not bisected anything.

Solution

  1. AB, BC and AC are all one radius long.
  2. So triangle ABC is equilateral.
  3. Every angle in an equilateral triangle is 60°.

Answer: D. 60°.

Correct.

Solution

  1. AB, BC and AC are all one radius long.
  2. So triangle ABC is equilateral.
  3. Every angle in an equilateral triangle is 60°.

2.To construct triangle PQR, Sara has drawn PQ = 9 cm. PR = 7 cm and QR = 8 cm. Which arcs should she draw next?

Not quite. This swaps the lengths. PR starts at P, so the arc from P has radius 7 cm.

Solution

  1. R is 7 cm from P (PR = 7 cm), so draw an arc of radius 7 cm from P.
  2. R is 8 cm from Q (QR = 8 cm), so draw an arc of radius 8 cm from Q.
  3. R is where the two arcs cross.

Answer: B. An arc of radius 7 cm from P and an arc of radius 8 cm from Q.

Correct.

Solution

  1. R is 7 cm from P (PR = 7 cm), so draw an arc of radius 7 cm from P.
  2. R is 8 cm from Q (QR = 8 cm), so draw an arc of radius 8 cm from Q.
  3. R is where the two arcs cross.

Not quite. PQ is already drawn. The arcs use the other two sides.

Solution

  1. R is 7 cm from P (PR = 7 cm), so draw an arc of radius 7 cm from P.
  2. R is 8 cm from Q (QR = 8 cm), so draw an arc of radius 8 cm from Q.
  3. R is where the two arcs cross.

Answer: B. An arc of radius 7 cm from P and an arc of radius 8 cm from Q.

Not quite. 15 = 7 + 8, but R is found with one arc from P and one arc from Q.

Solution

  1. R is 7 cm from P (PR = 7 cm), so draw an arc of radius 7 cm from P.
  2. R is 8 cm from Q (QR = 8 cm), so draw an arc of radius 8 cm from Q.
  3. R is where the two arcs cross.

Answer: B. An arc of radius 7 cm from P and an arc of radius 8 cm from Q.

3.Which set of lengths can not be the three sides of a triangle?

Correct.

Solution

  1. The two shorter sides must add up to more than the longest side.
  2. 3 + 5 = 8, which is less than 9.
  3. So the arcs from the ends of the 9 cm side cannot meet.

Not quite. 6 + 7 = 13, which is more than 12, so this triangle can be made.

Solution

  1. The two shorter sides must add up to more than the longest side.
  2. 3 + 5 = 8, which is less than 9.
  3. So the arcs from the ends of the 9 cm side cannot meet.

Answer: A. 3 cm, 5 cm and 9 cm.

Not quite. 5 + 5 = 10, which is more than 9, so this triangle can be made.

Solution

  1. The two shorter sides must add up to more than the longest side.
  2. 3 + 5 = 8, which is less than 9.
  3. So the arcs from the ends of the 9 cm side cannot meet.

Answer: A. 3 cm, 5 cm and 9 cm.

Not quite. 4 + 8 = 12, which is more than 11, so this triangle can be made.

Solution

  1. The two shorter sides must add up to more than the longest side.
  2. 3 + 5 = 8, which is less than 9.
  3. So the arcs from the ends of the 9 cm side cannot meet.

Answer: A. 3 cm, 5 cm and 9 cm.

Exam standard (Several steps, like real exam questions.)

4.Ella wants to construct an angle of 75° at a point A on a line. Which plan works?

Not quite. This gives 60° ÷ 4 = 15°.

Solution

  1. The angle between the 60° arm and the 90° arm is 30°.
  2. Bisecting it gives 15° on each side.
  3. From the line: 60° + 15° = 75°.

Answer: C. Construct 60° and 90° at A, on the same side of the line, then bisect the angle between them.

Not quite. This gives 90° ÷ 4 = 22.5°.

Solution

  1. The angle between the 60° arm and the 90° arm is 30°.
  2. Bisecting it gives 15° on each side.
  3. From the line: 60° + 15° = 75°.

Answer: C. Construct 60° and 90° at A, on the same side of the line, then bisect the angle between them.

Correct.

Solution

  1. The angle between the 60° arm and the 90° arm is 30°.
  2. Bisecting it gives 15° on each side.
  3. From the line: 60° + 15° = 75°.

Not quite. This gives 45° + 22.5° = 67.5°.

Solution

  1. The angle between the 60° arm and the 90° arm is 30°.
  2. Bisecting it gives 15° on each side.
  3. From the line: 60° + 15° = 75°.

Answer: C. Construct 60° and 90° at A, on the same side of the line, then bisect the angle between them.

5.Ivan constructs triangle ABC with AB = 12 cm, angle BAC = 60° and AC = 12 cm. How long is BC?

Not quite. This halves AB.

Solution

  1. AB = AC, so the triangle is isosceles and its base angles are equal.
  2. Each base angle is (180° − 60°) ÷ 2 = 60°.
  3. All three angles are 60°: the triangle is equilateral, so BC = 12 cm.

Answer: C. 12 cm.

Not quite. This uses Pythagoras as if angle A were 90°.

Solution

  1. AB = AC, so the triangle is isosceles and its base angles are equal.
  2. Each base angle is (180° − 60°) ÷ 2 = 60°.
  3. All three angles are 60°: the triangle is equilateral, so BC = 12 cm.

Answer: C. 12 cm.

Correct.

Solution

  1. AB = AC, so the triangle is isosceles and its base angles are equal.
  2. Each base angle is (180° − 60°) ÷ 2 = 60°.
  3. All three angles are 60°: the triangle is equilateral, so BC = 12 cm.

Not quite. This is the height of the triangle, not BC.

Solution

  1. AB = AC, so the triangle is isosceles and its base angles are equal.
  2. Each base angle is (180° − 60°) ÷ 2 = 60°.
  3. All three angles are 60°: the triangle is equilateral, so BC = 12 cm.

Answer: C. 12 cm.

6.The table shows the facts given for four triangles. Which of them does not fix one single triangle?

TriangleFacts given
Triangle 1AB = 7 cm, BC = 4 cm, CA = 6 cm
Triangle 2AB = 8 cm, angle A = 40°, AC = 6 cm
Triangle 3AB = 9 cm, angle A = 35°, angle B = 75°
Triangle 4AB = 9 cm, BC = 7 cm, angle A = 42°

Not quite. Three sides (SSS) fix one triangle.

Solution

  1. SSS, SAS and ASA each fix one triangle.
  2. Triangle 4 gives two sides and an angle that is NOT between them: angle A is not between AB and BC.
  3. The arc of radius 7 cm from B can cut the line from A twice, so two different triangles fit.

Answer: D. Triangle 4.

Not quite. Two sides and the angle between them (SAS) fix one triangle.

Solution

  1. SSS, SAS and ASA each fix one triangle.
  2. Triangle 4 gives two sides and an angle that is NOT between them: angle A is not between AB and BC.
  3. The arc of radius 7 cm from B can cut the line from A twice, so two different triangles fit.

Answer: D. Triangle 4.

Not quite. Two angles and the side between them (ASA) fix one triangle.

Solution

  1. SSS, SAS and ASA each fix one triangle.
  2. Triangle 4 gives two sides and an angle that is NOT between them: angle A is not between AB and BC.
  3. The arc of radius 7 cm from B can cut the line from A twice, so two different triangles fit.

Answer: D. Triangle 4.

Correct.

Solution

  1. SSS, SAS and ASA each fix one triangle.
  2. Triangle 4 gives two sides and an angle that is NOT between them: angle A is not between AB and BC.
  3. The arc of radius 7 cm from B can cut the line from A twice, so two different triangles fit.

7.Triangle XYZ is constructed with XY = 7 cm, angle X = 52° and angle Y = 71°. Which side of the triangle is the longest?

Not quite. XY is opposite angle Z = 57°, which is not the largest angle.

Solution

  1. Angle Z = 180° − 52° − 71° = 57°.
  2. The largest angle is angle Y = 71°.
  3. The longest side is opposite the largest angle: XZ.

Answer: B. XZ.

Correct.

Solution

  1. Angle Z = 180° − 52° − 71° = 57°.
  2. The largest angle is angle Y = 71°.
  3. The longest side is opposite the largest angle: XZ.

Not quite. YZ is opposite angle X = 52°, the smallest angle.

Solution

  1. Angle Z = 180° − 52° − 71° = 57°.
  2. The largest angle is angle Y = 71°.
  3. The longest side is opposite the largest angle: XZ.

Answer: B. XZ.

Not quite. No two angles are equal (52°, 71° and 57°), so no two sides are equal.

Solution

  1. Angle Z = 180° − 52° − 71° = 57°.
  2. The largest angle is angle Y = 71°.
  3. The longest side is opposite the largest angle: XZ.

Answer: B. XZ.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A student may construct the angle of an equilateral triangle and a right angle, bisect any angle, and add or subtract the angles made. Which of these angles can not be made this way?

Correct.

Solution

  1. Every angle made this way is a whole number of 15° steps, or of halves of them (7.5°, 3.75°, …).
  2. 15°, 105° and 135° are 1, 7 and 9 steps of 15°.
  3. 40° ÷ 15° = 83. Halving the step only doubles the top number, so it is never whole: 40° cannot be made.

Not quite. This can be made: bisect 60° twice.

Solution

  1. Every angle made this way is a whole number of 15° steps, or of halves of them (7.5°, 3.75°, …).
  2. 15°, 105° and 135° are 1, 7 and 9 steps of 15°.
  3. 40° ÷ 15° = 83. Halving the step only doubles the top number, so it is never whole: 40° cannot be made.

Answer: A. 40°.

Not quite. This can be made: 60° + 45°.

Solution

  1. Every angle made this way is a whole number of 15° steps, or of halves of them (7.5°, 3.75°, …).
  2. 15°, 105° and 135° are 1, 7 and 9 steps of 15°.
  3. 40° ÷ 15° = 83. Halving the step only doubles the top number, so it is never whole: 40° cannot be made.

Answer: A. 40°.

Not quite. This can be made: 90° + 45°.

Solution

  1. Every angle made this way is a whole number of 15° steps, or of halves of them (7.5°, 3.75°, …).
  2. 15°, 105° and 135° are 1, 7 and 9 steps of 15°.
  3. 40° ÷ 15° = 83. Halving the step only doubles the top number, so it is never whole: 40° cannot be made.

Answer: A. 40°.

9.Triangle ABC has AB = 10 cm, angle BAC = 30° and BC = 6 cm. How many different triangles fit these facts?

Not quite. Two sides and an angle NOT between them do not always fix one triangle.

Solution

  1. Draw AB and the 30° line from A. C must lie on this line.
  2. A 30° right-angled triangle is half of an equilateral triangle, so B is 10 ÷ 2 = 5 cm from the line.
  3. The arc of radius 6 cm from B: 6 is more than 5, so it reaches the line; 6 is less than AB, so it cuts it twice beyond A.
  4. Two positions for C: two triangles.

Answer: B. 2.

Correct.

Solution

  1. Draw AB and the 30° line from A. C must lie on this line.
  2. A 30° right-angled triangle is half of an equilateral triangle, so B is 10 ÷ 2 = 5 cm from the line.
  3. The arc of radius 6 cm from B: 6 is more than 5, so it reaches the line; 6 is less than AB, so it cuts it twice beyond A.
  4. Two positions for C: two triangles.

Not quite. B is 5 cm from the 30° line and 6 cm is more than 5 cm, so the arc does reach the line.

Solution

  1. Draw AB and the 30° line from A. C must lie on this line.
  2. A 30° right-angled triangle is half of an equilateral triangle, so B is 10 ÷ 2 = 5 cm from the line.
  3. The arc of radius 6 cm from B: 6 is more than 5, so it reaches the line; 6 is less than AB, so it cuts it twice beyond A.
  4. Two positions for C: two triangles.

Answer: B. 2.

Not quite. C must be on the 30° line and 6 cm from B. Only two points do this.

Solution

  1. Draw AB and the 30° line from A. C must lie on this line.
  2. A 30° right-angled triangle is half of an equilateral triangle, so B is 10 ÷ 2 = 5 cm from the line.
  3. The arc of radius 6 cm from B: 6 is more than 5, so it reaches the line; 6 is less than AB, so it cuts it twice beyond A.
  4. Two positions for C: two triangles.

Answer: B. 2.

10.An isosceles triangle is constructed by drawing its base, then arcs of radius 17 cm from each end of the base. The arcs cross 15 cm above the base. How long is the base?

Not quite. This is half of the base. The height meets the base at its midpoint, so double it.

Solution

  1. The height meets the base at its midpoint, making two right-angled triangles.
  2. Half-base² = 17² − 15² = 289 − 225 = 64, so half the base is 8 cm.
  3. The base is 2 × 8 = 16 cm.

Answer: C. 16 cm.

Not quite. This adds the squares. The 17 cm side is the hypotenuse, so subtract.

Solution

  1. The height meets the base at its midpoint, making two right-angled triangles.
  2. Half-base² = 17² − 15² = 289 − 225 = 64, so half the base is 8 cm.
  3. The base is 2 × 8 = 16 cm.

Answer: C. 16 cm.

Correct.

Solution

  1. The height meets the base at its midpoint, making two right-angled triangles.
  2. Half-base² = 17² − 15² = 289 − 225 = 64, so half the base is 8 cm.
  3. The base is 2 × 8 = 16 cm.

Not quite. This subtracts the lengths: 2 × (17 − 15). Use Pythagoras.

Solution

  1. The height meets the base at its midpoint, making two right-angled triangles.
  2. Half-base² = 17² − 15² = 289 − 225 = 64, so half the base is 8 cm.
  3. The base is 2 × 8 = 16 cm.

Answer: C. 16 cm.

Apply

Build it exactly

Use a ruler, a pencil and compasses (a protractor only for checking). Then compare with the model.

1. Construct triangle XYZ with XY = 7.6 cm, YZ = 6.3 cm and XZ = 5.4 cm.

2. Construct an angle of 135° at a point on a straight line.

3. Explain why no triangle has sides of 3.8 cm, 4.1 cm and 8.2 cm.

  1. Draw XY = 7.6 cm.
  2. Arc of radius 5.4 cm from X (XZ) and arc of radius 6.3 cm from Y (YZ). They cross at Z.
  3. Join XZ and YZ. Check each side to within 2 mm.
  1. Construct the perpendicular at the point: this gives two right angles.
  2. Bisect one of the right angles to make 45°.
  3. The angle from the line, past the 90° arm, to the 45° bisector of the other right angle is 90° + 45° = 135°.
  1. 3.8 + 4.1 = 7.9, which is less than 8.2.
  2. The arcs from the ends of the 8.2 cm side cannot meet, so there is no triangle.

Next: the Lesson 2 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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