57.5º Solution: As BC = AC, therefore triangle ABC is an isoscelestriangle. Given $\angle\text{ACD} = 115^\circ,\ \angle\text{ACB} = 180-115=65^\circ$ (Linear Pair) As AC = BC, therefore $\angle\text{A} =\angle\text{B}$ As sum of all the three angles of atriangle is 180° Therefore, $\angle\text{A} + \angle\text{B} + \angle\text{ACB} = 180^\circ$ $\angle\text{A} = \angle\text{B} = 57.5$
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