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In the figure, ABCD is a parallelogram. The measures of the angles $$x$$, $$y$$ and $$z$$ are
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The exterior angle at B is $$70^\circ$$, so $$\angle ABC = 110^\circ$$ and $$\angle DAB = 180^\circ - 110^\circ = 70^\circ$$. Then $$z = 70^\circ - 40^\circ = 30^\circ$$, and opposite angles give $$x = \angle ABC = 110^\circ$$. Since $$AD \parallel BC$$, $$y = \angle ACB = \angle CAD = 40^\circ$$ (alternate angles).
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