A) \[54{}^\circ \]
B) \[18{}^\circ \]
C) \[36{}^\circ \]
D) \[20{}^\circ \]
Correct Answer: C
Solution :
Here, J is the incentre of the \[\Delta ABC\] \[\therefore \]BI and CI are the bisectors of\[\angle B\]and \[\angle C\]then We know that, \[\angle BIC=90{}^\circ +\frac{1}{2}\angle A\]or \[108{}^\circ =90{}^\circ +\frac{1}{2}\angle A\] \[\Rightarrow \] \[\frac{1}{2}\angle A=108{}^\circ -90{}^\circ =18{}^\circ \] \[\angle A=36{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec
