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  1. To construct the required △ABC, here we use two facts on triangles: one is that, in a equilateral triangle, all three sides and three included angles are equal to each other, and the other is that, in a isoscelese triangle, the altitude bisects its base perpendicularly and also bisects the opposite vertex held by two sides of equal length.
  2. Mark a point B. From B, draw a ray rightwards, then an arc with radius of given length, BC=7.5cm, crossing the ray at C.  We are to construct vertex A above BC.  From B and C, draw 2 arcs with the same radius BC, crossing each other above BC at D.  From B thru D, draw a ray upwareds.  △DBC is an equilateral triangle.  ∠DBC=60°  Point A will be on the ray.
  3. From B, draw an arc with radius of given length, AB-AC=1.5cm, crossing BD at E.  Conect C to E.  Here we draw a perpendicular bisector of CE.  From C and E, draw 2 arcs with the same radius CE, crossing each othere above and below CE.  Draw a line thru the 2 crossings, intersecting the ray from B at A.  Connect A to C.  △AEC is an isoscelese triangle.
  4. CK: From A, draw an arc with radius AE.  The arc crosses AC at C. AE=AC, so AB-AC=AB-AE=BE=1.5cm.  CKD.  △ABC is the required triangle with ∠ABC=60°, BC=7.5cm and (AB-AC)=1.5cm.  (AB=approx.12.0 cm, AC=approx. 10.5cm).  

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