In the adjoining figure of \[\angle ABC,\text{ }\angle BCA=120{}^\circ \] and \[AB=c,\text{ }BC=a,\text{ }AC=b\] then:

\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-ba\]

\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}+ba\]

\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ba\]

\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}+2ba\]

In the adjoining figure of \[\angle ABC,\text{ }\angle BCA=120{}^\circ

1.	In the adjoining figure of \[\angle ABC,\text{ }\angle BCA=120{}^\circ \] and \[AB=c,\text{ }BC=a,\text{ }AC=b\] then:
A.	\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}+ba\]
B.	\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-ba\]
C.	\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ba\]
D.	\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}+2ba\]
E.	None of these
Answer» B. \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-ba\]

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In the adjoining figure of \[\angle ABC,\text{ }\angle BCA=120{}^\circ \] and \[AB=c,\text{ }BC=a,\text{ }AC=b\] then:

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