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Answer :
In \(\triangle{ABC}\), we have,
AB = AC ...(Given)
\(\angle{B}\) = \(\angle{C}\) ...(i)(\(\because \) angles opposite to equal sides are equal)
Now, we know that,
\(\angle{A}\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\)
\(\Rightarrow \) \(90^\circ\) + \(\angle{B}\) + \(\angle{C}\) = \(180^\circ\) ...(given)
\(\Rightarrow \) \(90^\circ\) + \(\angle{B}\) + \(\angle{B}\) = \(180^\circ\) ...(from(i))
\(\Rightarrow \) 2 \(\angle{B}\) = \(180^\circ\) - \(90^\circ\)
\(\Rightarrow \) 2 \(\angle{B}\) = \(90^\circ\)
\(\Rightarrow \) \(\angle{B}\) = \(45^\circ\)
\(\therefore \) \(\angle{C}\) = \(45^\circ\), too.