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Mark a point $$A$$, where the perpendicular bisector intersects the ray $$BX$$. Then, join $$AC$$.
Draw a line segment $$BC = 8 \ cm$$.
Draw a perpendicular bisector $$EF$$ of line segment $$CD$$.
The line $$BX$$ extended to opposite side of the line segment $$BC$$. With $$B$$ as the centre and $$2.5 \ cm$$ as radius, draw an arc that cuts the extended $$BX$$ at $$D$$. Then, join $$CD$$.
Make $$\angle XBC = 60^\circ$$.
Construct a triangle $$ABC$$ in which $$BC = 8 \ cm$$, $$\angle B = 60^\circ$$ and $$AC - AB = 60 \ cm$$.

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Thus, $$ABC$$ is the required triangle.