In figure $ \angle ABC = {69^0},\,\,\angle ACB = {31^0},\,\,find\,\,\angle BDC. $

Hint: Here, to find the value of $ \angle BDC $ we first apply angle sum property of triangle to get third angle of triangle as its two angles are given and then using the concept of circle that angles subtended by chord on arc of a circle are equal.

Complete step-by-step answer:
Here, it is given that $ \angle ABC = {69^0},\,\,\angle ACB = {31^0} $
Now, in triangle ABC, by angle sum property of triangle we have
 $ \angle BAC + \angle ABC + \angle ACB = {180^0} $
Substituting values in above equation. We have,
 $
  \angle BAC + {69^0} + {31^0} = {180^0} \\
   \Rightarrow \angle BAC + {100^0} = {180^0} \\
   \Rightarrow \angle BAC = {180^0} - {100^0} \\
   \Rightarrow \angle BAC = {80^0} \;
  $
Also, we know that in a circle angle subtended by a chord on the same arc of a circle are equal.
Hence, from the figure we see that $ \angle BAC\,\,and\,\,\angle BDC $ are angles subtended by chord BC on the arc of a circle.
 $ \therefore \angle BAC = \angle BDC $
But, the value of $ \angle BAC\,\,is\,\,{80^0}. $ Calculated above. Therefore using this in above we have,
 $ \angle BDC = {80^0} $
Hence, from above we see that the required value of $ \angle BDC\,\,is\,\,{80^0}. $

Note: From the figure we see that chord BC subtends two angles on the arc of the circle. Let the value of $ \angle BDC = x $ , then as $ \angle BDC = \angle BAC $ implies that value of $ \angle BAC = x $ . And, now in triangle ABC by using angle sum property of triangle and using it to form an equation by taking sum of all three angles equal to $ {180^0} $ and then solving the equation so formed to find value of ‘x’ and hence required value of $ \angle BDC. $