A) \[90{}^\circ -\frac{A}{A}\]
B) \[90{}^\circ +\frac{A}{A}\]
C) \[90{}^\circ -\frac{A}{2}\]
D) \[90{}^\circ +\frac{A}{2}\]
Correct Answer: D
Solution :
[d] Now, in \[\Delta \Beta \Iota C\] \[x{}^\circ +\frac{B}{2}+\frac{C}{2}=180{}^\circ \] \[\Rightarrow \] \[x{}^\circ =180{}^\circ -\frac{1}{2}(B+C)\] \[\Rightarrow \] \[x{}^\circ =180{}^\circ -\frac{1}{2}(180{}^\circ -A)\] \[\Rightarrow \] \[x{}^\circ =180{}^\circ -90{}^\circ +\frac{A}{2}\] \[\therefore \] \[\angle BIC=x{}^\circ -90{}^\circ +\frac{A}{2}\] |
You need to login to perform this action.
You will be redirected in
3 sec