
In the given figure, $\angle ABC = 69^\circ ,\angle ACB = 31^\circ . $ Find $ \angle BDC $
Answer
568.5k+ views
Hint: Apply angle sum property of triangle. And then use the properties of the circle to find $ \angle BDC $ ,Always remember the sum of angles in any triangle is 180 degree.
Complete step-by-step answer:
It is given in the question that,
$\angle ABC = 69^\circ $ and $ \angle ACB = 31^\circ $
We know that, in any triangle, sum of all the angles is equal to $ {180^0} $
Therefore, in $ \Delta ABC $
$ \angle BAC + \angle ABC + \angle ACB = 180^\circ $ (angle sum property of triangle)
$
\angle BAC + 69^\circ + 31^\circ = 180^\circ \\
\angle BAC + 100^\circ = 180^\circ \\
\angle BAC = 80^\circ \\
$
In any circle, angles opposite to the same arc are equal.
From the diagram, we can observe that,
$ \angle BDC $ and $ \angle BAC $ are angles opposite to the same arc, $ arc(BC) $
Therefore, $ \angle BDC = \angle BAC $ (Angles subtended by the same segment are equal)
$ \Rightarrow \angle BDC = 80^\circ $
Note: When angles are congruent, they have exactly the same in measure.
The angle subtended at the center by some arc is twice the angle subtended on the circumference of the circle by the same arc.
Complete step-by-step answer:
It is given in the question that,
$\angle ABC = 69^\circ $ and $ \angle ACB = 31^\circ $
We know that, in any triangle, sum of all the angles is equal to $ {180^0} $
Therefore, in $ \Delta ABC $
$ \angle BAC + \angle ABC + \angle ACB = 180^\circ $ (angle sum property of triangle)
$
\angle BAC + 69^\circ + 31^\circ = 180^\circ \\
\angle BAC + 100^\circ = 180^\circ \\
\angle BAC = 80^\circ \\
$
In any circle, angles opposite to the same arc are equal.
From the diagram, we can observe that,
$ \angle BDC $ and $ \angle BAC $ are angles opposite to the same arc, $ arc(BC) $
Therefore, $ \angle BDC = \angle BAC $ (Angles subtended by the same segment are equal)
$ \Rightarrow \angle BDC = 80^\circ $
Note: When angles are congruent, they have exactly the same in measure.
The angle subtended at the center by some arc is twice the angle subtended on the circumference of the circle by the same arc.
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