In the given figure the value of z is 120∘ 85∘ 80∘ 140∘

SOLUTION

In the figure, $\angle$ BCA and $\angle$ ACE are linear pair.
$\angle$ BCA + $\angle$ ACE $={180}^{\circ }$
            ${120}^{\circ }+x={180}^{\circ }\Rightarrow x={60}^{\circ }$

From $\Delta$ DCE, we have
${60}^{\circ }+{40}^{\circ }+y^{\circ }={180}^{\circ }$ 
                  [Angle sum property]
                   $\Rightarrow y={80}^{\circ }$

From exterior angle property of a triangle, we have
$x+y=z\Rightarrow z={60}^{\circ }+{80}^{\circ }={140}^{\circ }$
Hence, the value of 'z' is ${140}^{\circ }$