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Answer :

As we know that, the angles in the same segment are equal.

We get,

\(\angle{BDC}\) = \(\angle{BAC}\) ...(i)

Now, in \(\triangle{ABC}\),

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

\(\Rightarrow \) \(\angle{BAC}\) + \(69^\circ\) + \(31^\circ\) = \(180^\circ\)

\(\Rightarrow \)\(\angle{ABC}\) + \(100^\circ\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{ABC}\) = \(180^\circ\) - \(100^\circ\)

\(\therefore \) \(\angle{BAC}\) = \(80^\circ\)

\(\therefore \) \(\angle{BDC}\) = \(80^\circ\) ...(from (i))

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