A) \[\frac{3\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{2\pi }{3}\]
D) \[\frac{5\pi }{6}\]
Correct Answer: C
Solution :
Let \[a=\alpha -\beta ,b=\alpha +\beta ,c=\sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}}\] \ \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\] Þ \[\cos C=\frac{{{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -3{{\alpha }^{2}}-{{\beta }^{2}}}{2({{\alpha }^{2}}-{{\beta }^{2}})}\] Þ \[\cos C=-\frac{({{\alpha }^{2}}-{{\beta }^{2}})}{2({{\alpha }^{2}}-{{\beta }^{2}})}=\cos \left( \frac{2\pi }{3} \right)\] Þ \[\angle C=\frac{2\pi }{3}\], (largest angle).
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