An isosceles triangle \( A B C \) in which \( A B=B C=6 \sqrt{2} \) and \( A C=12 \) is folded along the altitude \( B D \) so the planes \( A B D \) and \( B D C \) form a right dihedral angle. Find the angle between side \( A B \) and its new position.
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Since \(ABC\) is an isosceles triangle, \(BD\) is also a median, so by the Pythagorean theorem, \(BD = \sqrt{(AB)^2 - (AC/2)^2} = \sqrt{(6\sqrt{2})^2 - (12/2)^2} = 6\). Show more…
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