| In a right angled triangle, the product of two sides is equal to half of the square of the third side, i.e. hypotenuse. One of the acute angles must be [SSC (Multitasking) 2014] |
A) \[60{}^\circ \]
B) \[30{}^\circ \]
C) \[45{}^\circ \]
D) \[15{}^\circ \]
Correct Answer: C
Solution :
| According to the question, |
| \[AB\cdot BC=\frac{1}{2}\cdot A{{C}^{2}}\] |
 |
| \[\Rightarrow \] \[2AB\cdot BC=A{{C}^{2}}\] |
| \[\Rightarrow \] \[2AB\cdot BC=A{{B}^{2}}+B{{C}^{2}}\] |
| \[\Rightarrow \]\[A{{B}^{2}}+B{{C}^{2}}-2AB\cdot BC=0\] |
| \[\Rightarrow \] \[{{(AB-BC)}^{2}}=0\] |
| \[\Rightarrow \] \[AB=BC\] |
| \[\therefore \] \[\angle BAC=\angle BCA=45{}^\circ \] |
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