question_answer

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.