A) \[45{}^\circ \]
B) \[60{}^\circ \]
C) \[90{}^\circ \]
D) \[120{}^\circ \]
Correct Answer: D
Solution :
\[{{C}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \] Since, \[A=B=C\] \[{{A}^{2}}={{A}^{2}}+{{A}^{2}}+2A\times Acos\theta \] \[{{A}^{2}}=2{{A}^{2}}(1+cos\theta )\] \[1+\cos \theta =\frac{1}{2}\] \[\cos \theta =\frac{1}{2}-1=-\frac{1}{2}=\cos {{120}^{o}}\] \[\theta ={{120}^{o}}\]You need to login to perform this action.
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