If I is the in centre of \[\Delta ABC\] and \[\angle BIC=135{}^\circ ,\] then \[\Delta ABC\] is |
A) acute angled
B) equilateral
C) right angled
D) obtuse angled
Correct Answer: C
Solution :
\[\because \angle BIC=135{}^\circ \] |
In \[\Delta BIC,\] |
\[\angle BIC+\angle IBC+\angle ICB=180{}^\circ \] |
\[\Rightarrow \]\[\angle IBC+\angle ICB=180{}^\circ -135{}^\circ =45{}^\circ \] |
\[\Rightarrow \] \[2\angle IBC+2\angle ICB=90{}^\circ \] |
\[\Rightarrow \] \[\angle B+\angle C=90{}^\circ \] |
Now, \[\angle A+\angle B+\angle C=180{}^\circ \] |
\[\therefore \] \[\angle A=90{}^\circ \] |
Hence, \[\Delta ABC\] is a right angled triangle. |
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