The internal bisectors of ∠B and ∠C of △ABC meet at D. If ∠A= $$75^∘$$​, the ∠BDC is:

Correct option is C

Given :​

$△ABC$ with ∠A=75∘\angle A = 75^\circ∠ A =75∘.

Internal angle bisectors of $\angle B$ and $\angle C$ meet at $D$.

​to find $\angle BDC$.

Formula used :

Using the property of internal angle bisectors : 

$$\angle BDC = 90^\circ+\frac{\angle A}{2}$$ 

Solution:

Using the property;

 $$\angle BDC = 90^\circ+\frac{75^\circ}{2}=127.5$$​

The correct answer is option ( c ) 127.5

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