A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[90{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: C
Solution :
[c] We have, \[AC=BC\] |
and \[A{{B}^{2}}=2A{{C}^{2}}=A{{C}^{2}}+A{{C}^{2}}\] |
\[\Rightarrow \,\,\,\,\,\,\,\,A{{B}^{2}}=B{{C}^{2}}+A{{C}^{2}}\] \[[AC=BC]\] |
\[\therefore \]By converse of Pythagoras theorem. \[\angle C=90{}^\circ \] |
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