4. In figure, $$\angle{ABC}$$ = $$69^\circ$$, $$\angle{ACB}$$ = $$31^\circ$$. Find $$\angle{BDC}$$.

We get, $$\angle{BDC}$$ = $$\angle{BAC}$$ ...(i)
Now, in $$\triangle{ABC}$$,
$$\angle{BAC}$$ + $$\angle{ABC}$$ + $$\angle{ACB}$$ = $$180^\circ$$
i.e., $$\angle{BAC}$$ + $$69^\circ$$ + $$31^\circ$$ = $$180^\circ$$
i.e., $$\angle{ABC}$$ + $$100^\circ$$ = $$180^\circ$$
i.e., $$\angle{ABC}$$ = $$180^\circ$$ - $$100^\circ$$
So, $$\angle{BAC}$$ = $$80^\circ$$
Therefore, $$\angle{BDC}$$ = $$80^\circ$$ ...(from (i))