Draw a line XY.
Mark a point D on XY and construct DZ perpendicular to XY.
With center D and radius 3.2cm cut an arc on DZ at A.
With A as center, draw ∠UAD=30 intersecting XY at B and ∠VAD=30° intersecting XY at C.
ABC is the required triangle.
By construction we can say,
∠BAC = ∠BAD + ∠CAD
∠BAC = 30°+30° =60°.
∠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.
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