Construction: 60° Angle
Constructing a 60° angle using only a compass and straightedge (ruler without markings) is one of the fundamental constructions in Class 9 geometry. This construction relies on the property of an equilateral triangle, where all three angles measure 60°.
The 60° angle construction is a building block for constructing other standard angles. By bisecting a 60° angle, a 30° angle is obtained. By adding a 60° angle to itself, a 120° angle is formed. A 90° angle can be constructed by bisecting the 120° angle supplementary to the 60° angle.
In CBSE Class 9, students learn to perform constructions using compass and straightedge only — no protractor is allowed. The justification for each construction step is based on geometric properties and theorems.
What is Construction: 60° Angle?
Definition: A geometric construction is a drawing made using only two instruments: a compass (for drawing arcs and circles) and a straightedge (for drawing straight lines, without measurement).
A 60° angle is constructed using the property of an equilateral triangle
Why It Works:
- An equilateral triangle has all sides equal and all angles equal to 60°.
- When two arcs of the same radius are drawn from the endpoints of a line segment and they intersect, the triangle formed is equilateral.
- Therefore, the angle at the base is exactly 60°.
Important:
- The compass width must NOT be changed between drawing the two arcs.
- A protractor is NOT allowed in constructions — only compass and straightedge.
- The construction must be accompanied by a justification explaining why the angle is 60°.
Construction: 60° Angle Formula
Key Construction Steps and Derived Angles:
Angles that can be constructed from 60°:
- 60° — Direct construction (equilateral triangle property)
- 30° — Bisect the 60° angle
- 15° — Bisect the 30° angle
- 120° — Construct 60°, then construct another 60° adjacent to it
- 90° — Bisect 120° (or construct perpendicular bisector)
- 45° — Bisect 90°
- 75° — Construct 60° + bisect the adjacent angle to get 15° more
- 150° — 180° − 30° (construct 30° and take supplement)
Key Principle:
Equal radii from the same point create equal lengths → Equilateral triangle → 60°
Derivation and Proof
Step-by-Step Construction of a 60° Angle:
Given: A ray OA.
To construct: An angle of 60° at O.
Steps of Construction:
- Draw a ray OA.
- With O as centre and any convenient radius r, draw an arc that intersects ray OA at a point, say P.
- With P as centre and the same radius r, draw another arc that intersects the first arc at a point, say Q.
- Draw ray OQ.
- ∠QOP = 60°.
Justification:
- OP = r (radius of the first arc)
- PQ = r (radius of the second arc, same as the first)
- OQ = r (OQ is also a radius of the first arc, since Q lies on it)
- Therefore, △OPQ is equilateral (all three sides = r).
- All angles of an equilateral triangle = 60°.
- Therefore, ∠QOP = 60°.
Construction of 30° (by bisecting 60°):
- Construct 60° angle QOP as above.
- With P as centre and radius more than half of PQ, draw an arc.
- With Q as centre and the same radius, draw another arc intersecting the first at point R.
- Draw ray OR.
- ∠POR = ∠ROQ = 30° (OR bisects the 60° angle).
Construction of 120°:
- Construct the 60° angle QOP as above.
- With Q as centre and the same radius r, draw another arc intersecting the first arc at point S (on the other side of Q from P).
- Draw ray OS.
- ∠POS = ∠POQ + ∠QOS = 60° + 60° = 120°.
Types and Properties
Types of Angle Constructions Using 60° as Base:
1. Direct 60° Construction
- Uses the equilateral triangle property.
- Requires only 2 arcs (one from O, one from the intersection point on the ray).
- Simplest of all angle constructions.
2. Construction of 30°
- First construct 60°, then bisect it.
- The bisector is found using two arcs of equal radius from the arms of the 60° angle.
- Total arcs needed: 2 (for 60°) + 2 (for bisector) = 4 arcs.
3. Construction of 120°
- Construct 60°, then construct another 60° adjacent to it.
- Uses 3 arcs: the base arc, first intersection (60°), second intersection (120°).
4. Construction of 90°
- Method 1: Construct 60° and 120°, then bisect the angle between 60° and 120°.
- Method 2: Use perpendicular bisector construction (separate method).
5. Construction of 150°
- Construct 30° on a straight line.
- The supplement of 30° is 150°.
6. Construction of 45°
- Construct 90° and bisect it.
- Or construct 60°, bisect to get 30°, construct 60° from the other arm and bisect the resulting angle.
Solved Examples
Example 1: Example 1: Construct a 60° angle at point O
Problem: Using compass and straightedge, construct an angle of 60° at a given point O on a ray OA.
Construction:
- Draw ray OA.
- With O as centre and radius 5 cm, draw an arc cutting OA at P.
- With P as centre and the same radius (5 cm), draw an arc cutting the first arc at Q.
- Join OQ and extend it to form ray OB.
Result: ∠BOA = 60°
Justification: OP = PQ = OQ = 5 cm. Triangle OPQ is equilateral. Therefore ∠POQ = 60°.
Example 2: Example 2: Construct a 30° angle
Problem: Construct an angle of 30° using compass and straightedge.
Construction:
- First construct a 60° angle: Draw ray OA. With centre O and any radius, draw an arc cutting OA at P. With centre P and same radius, draw arc cutting first arc at Q. Join OQ to form ∠QOP = 60°.
- Now bisect ∠QOP: With centre P and radius more than half PQ, draw an arc. With centre Q and same radius, draw arc intersecting at R.
- Join OR. ∠POR = 30°.
Justification: OR is the perpendicular bisector of PQ (by construction). Since OP = OQ (radii) and PR = QR, OR bisects ∠POQ. Therefore ∠POR = 60°/2 = 30°.
Example 3: Example 3: Construct a 120° angle
Problem: Construct an angle of 120° using compass and straightedge.
Construction:
- Draw ray OA. With centre O and any radius, draw a large arc cutting OA at P.
- With centre P and same radius, cut the arc at Q. (∠POQ = 60°)
- With centre Q and same radius, cut the arc at S. (∠QOS = 60°)
- Join OS to form ray OB. ∠POB = ∠POQ + ∠QOS = 60° + 60° = 120°.
Result: ∠BOA = 120°.
Example 4: Example 4: Construct a 90° angle
Problem: Construct a 90° angle using compass and straightedge.
Construction:
- Construct 60° and 120° as in Examples 1 and 3 above. Let Q be the 60° mark and S be the 120° mark on the arc.
- Now bisect ∠QOS (which is 60°): With centre Q and radius more than half QS, draw an arc. With centre S and same radius, draw arc cutting at T.
- Join OT. ∠POT = 60° + 30° = 90°.
Justification: OT bisects the 60° angle between 60° and 120°, giving 60° + 30° = 90° from the ray OA.
Example 5: Example 5: Construct a 150° angle
Problem: Construct an angle of 150°.
Construction:
- Draw a straight line and mark a point O on it. Call the line XOA (X on the left, A on the right).
- Construct 30° on ray OA (construct 60° and bisect it). Let this give ray OB where ∠BOA = 30°.
- ∠BOX = 180° − 30° = 150°.
Result: ∠BOX = 150°.
Example 6: Example 6: Construct an equilateral triangle with side 5 cm
Problem: Construct an equilateral triangle ABC with side 5 cm using the 60° angle construction.
Construction:
- Draw segment AB = 5 cm.
- At point A, construct a 60° angle: With A as centre and radius 5 cm, draw an arc. With the point where the arc meets AB as centre and same radius, draw an arc meeting the first arc at C.
- Join AC and BC.
Result: △ABC is equilateral with AB = BC = CA = 5 cm and all angles = 60°.
Example 7: Example 7: Construct 15°
Problem: Construct an angle of 15°.
Construction:
- Construct 60° at point O on ray OA.
- Bisect the 60° to get 30°.
- Bisect the 30° angle to get 15°.
Justification: 60° ÷ 2 = 30°. Then 30° ÷ 2 = 15°. Each bisection is done using the standard compass method.
Example 8: Example 8: Construct 75°
Problem: Construct an angle of 75°.
Construction:
- Construct 60° at O on ray OA. Call this ray OB.
- Construct 90° at O on ray OA. Call this ray OC.
- ∠BOC = 90° − 60° = 30°
- Bisect ∠BOC to get a ray OD such that ∠BOD = 15°.
- ∠DOA = 60° + 15° = 75°.
Result: ∠DOA = 75°.
Example 9: Example 9: Construct an isosceles triangle with vertex angle 120°
Problem: Construct an isosceles triangle PQR with PQ = PR = 4 cm and ∠QPR = 120°.
Construction:
- Draw a ray PA.
- Construct 120° at P (using the two successive 60° arcs method).
- On ray PA, mark Q at 4 cm from P.
- On the other arm of the 120° angle, mark R at 4 cm from P.
- Join QR.
Result: △PQR with PQ = PR = 4 cm and ∠QPR = 120°.
Example 10: Example 10: Verify the construction with a protractor
Problem: Construct a 60° angle and verify using a protractor.
Construction:
- Draw ray OA.
- With O as centre and radius 4 cm, draw an arc cutting OA at P.
- With P as centre and radius 4 cm, draw an arc cutting the first arc at Q.
- Join OQ to form ray OB.
Verification:
- Place the protractor with centre at O and baseline along OA.
- Read the angle where ray OB crosses the protractor scale.
- The reading should be 60°.
Answer: The constructed angle measures 60°, confirmed by protractor verification.
Real-World Applications
Applications of 60° Angle Construction:
- Constructing Other Standard Angles: The 60° construction is the starting point for 30°, 15°, 120°, 90°, 45°, 75°, and 150° constructions. Nearly all constructible angles are derived from 60° and its multiples.
- Constructing Regular Polygons: A regular hexagon has internal angles of 120° and central angles of 60°. The 60° construction enables drawing regular hexagons, which are common in tiling and honeycomb patterns.
- Engineering Drawing: Technical drawings in engineering use compass-and-straightedge constructions for precision. The 60° angle appears in bolt-head hexagons, threaded fasteners, and truss designs.
- Architecture: Triangular roof trusses often use 60° angles for optimal load distribution. The equilateral triangle construction is used in designing these structures.
- Art and Design: The 60° angle creates visually balanced compositions. It is used in geometric art, Islamic tessellation patterns, and Celtic knot designs.
- Navigation: Bearing calculations sometimes require constructing precise angles using compass methods, especially in traditional navigation without electronic instruments.
Key Points to Remember
- A 60° angle is constructed using the property that an equilateral triangle has all angles equal to 60°.
- The construction requires only a compass and straightedge — no protractor.
- The key step: draw an arc from a point on the ray with the same radius as the arc from the vertex. The intersection gives the 60° point.
- The compass width must not change between the two arcs.
- 30° = bisect 60°. 120° = two successive 60° arcs. 90° = bisect between 60° and 120°.
- Every construction must include a justification (why the angle has the claimed measure).
- The justification for 60° uses: three equal radii → equilateral triangle → 60°.
- Construction of 15° requires two successive bisections (60° → 30° → 15°).
- Construction of 75° = 60° + half of 30° (or bisect the gap between 60° and 90°).
- The constructed angle should always be verified with a protractor in practice problems.
Practice Problems
- Construct a 60° angle at a point on a given line. Write the justification.
- Construct a 30° angle using only compass and straightedge. List all steps.
- Construct an angle of 120° and verify with a protractor.
- Construct a 90° angle using the 60° and 120° construction method.
- Construct a 45° angle starting from a 90° construction.
- Construct an equilateral triangle with side 6 cm. Measure all angles to verify.
- Construct a 75° angle. Show all construction arcs and write the justification.
- Can you construct a 40° angle using compass and straightedge only? Justify your answer.
Frequently Asked Questions
Q1. How do you construct a 60° angle?
Draw a ray OA. With centre O and any radius, draw an arc cutting OA at P. With centre P and the same radius, draw an arc cutting the first arc at Q. Join OQ. The angle QOP is 60°.
Q2. Why does this construction give exactly 60°?
Because OP = PQ = OQ (all equal to the compass radius). This makes triangle OPQ equilateral. All angles of an equilateral triangle are 60°.
Q3. Can you use a protractor to construct 60°?
Yes, a protractor can measure and draw 60°. However, in CBSE Class 9 constructions, only compass and straightedge are allowed. The compass-straightedge method is exact, while protractor measurement may have small reading errors.
Q4. How do you construct 30° from 60°?
Construct 60° first. Then bisect it: from both points on the arms (where the arc cuts), draw equal arcs that intersect. The line from the vertex through this intersection bisects the 60° angle, giving 30°.
Q5. Which angles can be constructed with compass and straightedge?
Angles that are multiples of 15° can be constructed: 15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180°. Angles like 40°, 50°, or 70° cannot be constructed exactly.
Q6. Why can't you change the compass width between the two arcs?
If the compass width changes, OP, PQ, and OQ will not all be equal. The triangle OPQ will not be equilateral, and the angle will not be 60°.
Q7. How do you construct 120°?
Draw the first arc from O. Mark two successive points using the same radius: the first gives 60°, the second gives 120°. Join the vertex to the second point.
Q8. What is the justification for 30°?
The bisector of 60° creates two equal angles of 30° each. The bisector construction is justified by the fact that the two intersection arcs create a point equidistant from both arms, which lies on the angle bisector.
Q9. Is the 60° angle construction in the CBSE Class 9 syllabus?
Yes. Construction of 60° and derived angles (30°, 90°, 120°, 45°) is part of the CBSE Class 9 chapter on Constructions.
Q10. Can a 1° angle be constructed with compass and straightedge?
No. It is mathematically proven that a 1° angle cannot be constructed with compass and straightedge alone. Only angles that are multiples of 3° (and specifically involve factors of 2, 3, and certain Fermat primes) can be constructed.
