A) \[60{}^\circ ~\]
B) \[30{}^\circ \]
C) \[45{}^\circ \]
D) \[15{}^\circ \]
Correct Answer: C
Solution :
[c] According to the question, \[AB.BC=\frac{1}{2}.A{{C}^{2}}\] \[\Rightarrow \] \[2\,AB.\,\,BC=A{{C}^{2}}\] \[\Rightarrow \] \[2\,AB.\,\,\,BC=A{{B}^{2}}+B{{C}^{2}}\] \[\Rightarrow \] \[A{{B}^{2}}+B{{C}^{2}}-2AB.\,\,BC=0\] \[\Rightarrow \] \[{{(AB-BC)}^{2}}=0\]\[\Rightarrow \]\[AB=BC\] \[\therefore \] \[\angle BAC=\angle BCA=45{}^\circ \] |
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