Angles with a compass
Draw BC = 5 cm and construct ∠ABC = 60° at B.
Draw BC = 4.5 cm and construct ∠ABC = 120° at B.
Draw BC = 5 cm and construct ∠ABC = 90° at B.
Draw BC = 6 cm and construct ∠ABC = 30° at B.
Draw BC = 5.5 cm and construct ∠ABC = 45° at B.
1. Draw BC = 5 cm.
2. With B as the center and any radius (here 2 cm), draw an arc that cuts BC at P.
3. With P as the center and the same radius, cut the arc at Q.
4. Join BQ and produce it to A: ∠ABC = 60°.
1. Draw BC = 4.5 cm.
2. With B as the center and any radius (here 2 cm), draw an arc that cuts BC at P.
3. With P as the center and the same radius, cut the arc at Q.
4. With Q as the center and the same radius, cut the arc again at R.
5. Join BR and produce it to A: ∠ABC = 120°.
1. Draw BC = 5 cm.
2. With B as the center and any radius (here 2 cm), draw an arc that cuts BC at P.
3. With P as the center and the same radius, cut the arc at Q.
4. With Q as the center and the same radius, cut the arc again at R.
5. With Q and R as the centers and the same radius, draw arcs that cut at S.
6. Join BS and produce it to A: BS bisects ∠RBQ, so ∠ABC = 60° + 30° = 90°.
1. Draw BC = 6 cm.
2. With B as the center and any radius (here 2 cm), draw an arc that cuts BC at P.
3. With P as the center and the same radius, cut the arc at Q.
4. With P and Q as the centers and the same radius, draw arcs that cut at S.
5. Join BS and produce it to A: BS bisects ∠QBC = 60°, so ∠ABC = 30°.
1. Draw BC = 5.5 cm.
2. With B as the center and any radius (here 2 cm), draw an arc that cuts BC at P.
3. With P as the center and the same radius, cut the arc at Q.
4. With Q as the center and the same radius, cut the arc again at R.
5. With Q and R as the centers and the same radius, draw arcs that cut at S.
6. Join BS. It cuts the first arc at T, and ∠TBC = 90°.
7. With P and T as the centers and the same radius, draw arcs that cut at U.
8. Join BU and produce it to A: BU bisects ∠TBP, so ∠ABC = 0° + 45° = 45°.
Why it works
Every arc has the same radius, so B, P and Q are the corners of an equilateral triangle, and ∠QBP = 60°. Each further cut adds another 60°. Arcs of the same radius from two points on the arc meet on the line that bisects the angle between them: halving gives 30° from 60°, 90° between 60° and 120°, and so on.