Answer: The answer is (A). Similar.
Step-by-step explanation: We are given two triangles, ΔABC and ΔDEF where, ∠B = 80°, ∠C = 60°, ∠D = 40° and ∠E = 80°.
We are to check whether the two triangles are similar or not.
In ΔABC, we have
![\angle A+\angle B+\angle C=180^\circ\\\\\Rightarrow \angle A+80^\circ+60^\circ=180^\circ\\\\\Rightarrow \angle A=180^\circ-140^\circ\\\\\Rightarrow \angle A=40^\circ,](https://tex.z-dn.net/?f=%5Cangle%20A%2B%5Cangle%20B%2B%5Cangle%20C%3D180%5E%5Ccirc%5C%5C%5C%5C%5CRightarrow%20%5Cangle%20A%2B80%5E%5Ccirc%2B60%5E%5Ccirc%3D180%5E%5Ccirc%5C%5C%5C%5C%5CRightarrow%20%5Cangle%20A%3D180%5E%5Ccirc-140%5E%5Ccirc%5C%5C%5C%5C%5CRightarrow%20%5Cangle%20A%3D40%5E%5Ccirc%2C)
and in ΔDEF, we have
![\angle D+\angle E+\angle F=180^\circ\\\\\Rightarrow 40^\circ+80^\angle F=180^\circ\\\\\Rightarrow \angle F=180^\circ-120^\circ\\\\\Rightarrow \angle F=60^\circ.](https://tex.z-dn.net/?f=%5Cangle%20D%2B%5Cangle%20E%2B%5Cangle%20F%3D180%5E%5Ccirc%5C%5C%5C%5C%5CRightarrow%2040%5E%5Ccirc%2B80%5E%5Cangle%20F%3D180%5E%5Ccirc%5C%5C%5C%5C%5CRightarrow%20%5Cangle%20F%3D180%5E%5Ccirc-120%5E%5Ccirc%5C%5C%5C%5C%5CRightarrow%20%5Cangle%20F%3D60%5E%5Ccirc.)
Therefore, in ΔABC and ΔDEF, we have
∠A = ∠D,
∠B = ∠E,
∠C = ∠F.
So, by AAA similarity postulate, we get
ΔABC ≈ ΔDEF.
Thus, (A) is the correct option.