question_answer

In a \[\Delta ABC,\] \[\angle BCA=60{}^\circ \] and \[A{{B}^{2}}=B{{C}^{2}}+C{{A}^{2}}+X.\] What is the value X?

A) \[(BC)(CA)\]

B) \[-\left( BC \right)\,\,\left( AC \right)\]

C) \[(AB)(BC)\]

D) Zero

Correct Answer: B

Solution :

[b] By cosine law, \[\cos \,60{}^\circ =\frac{A{{C}^{2}}+B{{C}^{2}}-A{{B}^{2}}}{2.AC.BC}=\frac{1}{2}\] \[\Rightarrow \]   \[A{{C}^{2}}+B{{C}^{2}}-A{{B}^{2}}=AC\cdot BC\] \[\therefore \]By comparing, we get \[X=-\,(AC)(BC)\]