Let ABCD be a convex quadrilateral in which A | vedclass

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(C) In $	riangle ABC$ and $	riangle DCB$,$AB = DC$ (Given)$BC = CB$ (Common side)$AC = DB$ (Given)By $SSS$ congruence criterion,$	riangle ABC \cong

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$80$

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(C) In $	riangle ABC$ and $	riangle DCB$,$AB = DC$ (Given)$BC = CB$ (Common side)$AC = DB$ (Given)By $SSS$ congruence criterion,$	riangle ABC \cong 	riangle DCB$.Therefore,$\angle BAC = \angle CDB = 70^{\circ}$ and $\angle ABC = \angle DCB = 60^{\circ}$.In $	riangle ABC$,$\angle ACB = 180^{\circ} - \angle BAC - \angle ABC = 180^{\circ} - 70^{\circ} - 60^{\circ} = 50^{\circ}$.Now,$\angle DCO = \angle BCD - \angle ACB = 60^{\circ} - 50^{\circ} = 10^{\circ}$.In $	riangle DOC$,$\angle DOC = 180^{\circ} - \angle ODC - \angle DCO = 180^{\circ} - 70^{\circ} - 10^{\circ} = 100^{\circ}$.The acute angle between $AC$ and $BD$ is $180^{\circ} - 100^{\circ} = 80^{\circ}$.

Let $ABCD$ be a convex quadrilateral in which $AC = BD$,$AB = CD$,$\angle BAC = 70^{\circ}$ and $\angle BCD = 60^{\circ}$. The acute angle between $AC$ and $BD$ is (in $^{\circ}$)

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