In the given figure ΔABC, if θ = 80°, the measure of each of the other two angles will be:

50°

In isosceles triangle ABC

\(\angle\)ACB = \(\angle\)ABC [AC = AB]

Given, \(\angle\)BAC = \({80}^\circ\)

In triangle ABC

\(\angle\)ACB + \(\angle\)ABC + \(\angle\)BAC = \({180}^\circ\)

= 2\(\angle\)ACB = 180 - 80 = \({100}^\circ\)

= \(\angle\)ACB = \(\frac{100}{2}\) = \({50}^\circ\).