An equilateral triangle if its altitude is 3.2 cm.

Step 1:
Draw a line XY.

Step 2:

Mark a point D on XY and construct DZ perpendicular to XY.

Step 3:

With center D and radius 3.2cm cut an arc on DZ at A.

Step 4:

With A as center, draw ∠UAD=30 intersecting XY at B and ∠VAD=30° intersecting XY at C.

ABC is the required triangle.

Justification

By construction we can say,

∠ZDY=90°

∠BAC = ∠BAD + ∠CAD

∠BAC = 30°+30° =60°.

In ΔABD,

∠ABD + ∠BAD + ∠DBA = 180° (By Angle Sum Property)

30° + 90° + ∠DBA = 180°

∠DBA = 60°

Similary, ∠DCA = 60°

Thus, ∠A = ∠B = ∠C = 60°

Thus our construction is justified.