Construction of angles — 60°, 90°, 120°, 45°, 30° using compass only

Question

Construct a $30°$ angle using only a compass and straightedge. Also explain how to construct $60°$ , $90°$ , $120°$ , and $45°$ .

(CBSE 6 & 9 — constructions chapter)

Solution — Step by Step

This is the foundation for all other constructions.

Draw a ray $OA$
With centre $O$ , draw an arc of any radius cutting $OA$ at point $P$
With centre $P$ and same radius, cut the arc at point $Q$
Draw ray $OQ$ — angle $QOA = 60°$

Why 60°? The construction creates an equilateral triangle ( $OP = PQ = OQ$ ), and each angle of an equilateral triangle is $60°$ .

Construct $60°$ (from Step 1)
Bisect it: with centres $P$ and $Q$ (using same radius), draw arcs that intersect at point $R$
Draw ray $OR$ — this bisects the $60°$ angle
Angle $ROA = 30°$

120°: With centre $Q$ , cut the arc again at $S$ using the same radius. Angle $SOA = 120°$ (two copies of $60°$ ).
90°: Bisect the angle between $60°$ and $120°$ — that gives $90°$ .
45°: Bisect $90°$ .

Why This Works

All standard angle constructions trace back to the equilateral triangle ( $60°$ ). By bisecting and combining, we can reach any multiple of $15°$ .

graph TD
    A["Start: 60°<br/>(equilateral triangle)"] --> B["Bisect → 30°"]
    A --> C["Double → 120°"]
    A --> D["Bisect 60°-120° gap → 90°"]
    D --> E["Bisect 90° → 45°"]
    E --> F["Bisect 45° → 22.5°"]
    B --> G["Bisect 30° → 15°"]
    C --> H["Add 60° → 180°<br/>(straight line)"]
    D --> I["Bisect 90°-120° → 105°"]
    B --> J["Add 30° + 60° → 90°<br/>(alternative method)"]

The compass-and-straightedge limitation means we can only construct angles that are multiples of $3°$ (more precisely, those achievable through bisection and $60°$ combinations). Arbitrary angles like $1°$ or $20°$ cannot be constructed exactly with just these tools.

Alternative Method — 90° Without Using 60°

Construct $90°$ directly using a perpendicular:

Draw line $AB$
With $A$ as centre, draw arcs on both sides of $AB$
With the intersection points as centres, draw arcs that intersect at $C$
$CA \perp AB$ — this gives $90°$

For CBSE 9 boards: the examiner looks for construction arcs — DO NOT erase them. Leave all arcs and intersection points visible. Also label all points clearly. Marks are given for the construction steps, not just the final angle.

Common Mistake

When bisecting an angle, students sometimes change the compass radius between the two arcs (from the two arms to find the bisector). The radius must be the same for both arcs — otherwise the intersection point won’t lie on the actual bisector, and your angle will be slightly off. Keep the compass setting unchanged once you start the bisection step.

Question

Solution — Step by Step

Why This Works

Alternative Method — 90° Without Using 60°

Common Mistake

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