how to find the base of an isosceles triangle if the equal sides are given and the angle between them too.
for example... Q- find the base of the isosceles triangle if the equal sides are 6 cm each and the angle between them is 45 degree.
( try not to involve trigonometry )

You may use Pythagoras theorem



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With out use of sine or cosine of angles, it seems not possible to find it easily.

Let  ABC be the isosceles triangle, with A being the intersection of equal sides. So angle A = 45°.   angles B = C = (180°-45°)/2 = 67.5°.       
      Given   AB = AC = 6 cm.

The easy way is  : 
         AB / Sin C = BC / Sin A  = 2 R = circumcircle diameter

                =>      BC = AB * Sin A / Sin C  = 2 R * Sin A
        BC = base = 6 * Sin 45° / Sin 67.5°  = 4.592  cm