If in △ABC, AB = AC and ∠ACD = 125°, then ∠BAC is:

70°

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

= x + x + \(\angle\)BAC = 180

= \(\angle\)BAC = 180 - 2x

As, \(\angle\)ACD = exterior angle of triangle ABC

= \(\angle\)ACD = \(\angle\)BAC + \(\angle\)ACB

= 125 = 180 - 2x + x

= x = 55

\(\angle\)BAC = 180 - 2x = 180 - (2 x 55)

= 180 - 110 = 70

Therefore, \(\angle\)BAC is \({70}^\circ\).