In ΔABC, DE || BC in such a way that A-D-B and A-E-C. If m ∠ACB = 40°…

Here, we have a triangle ABC in which DE is parallel to BC

So, the \(\angle\)ACB = \(\angle\)AED = \({40}^\circ\)

Now, consider the triangle ADE as we know that

Sum of the angles of a triangle = 180

= \(\angle\)ADE + \(\angle\)DAE + \(\angle\)AED = 180

= \(\angle\)ADE + \(\angle\)DAE + 40 = 180

= \(\angle\)ADE + \(\angle\)DAE = \({140}^\circ\)

Therefore, the required angle is \({140}^\circ\).