A) A right angled triangle
B) An isosceles triangle
C) An equilateral triangle
D) None of these
Correct Answer: C
Solution :
\[{{S}_{1}}=\frac{1}{-2+\sqrt{4-1}}=\frac{1}{-2+\sqrt{3}}=-(\sqrt{3}+2)\] \[{{S}_{2}}=\frac{1}{-2-\sqrt{4-1}}=\frac{1}{-2-\sqrt{3}}=(\sqrt{3}-2)\] and \[{{S}_{3}}=1\]. \[{{\theta }_{13}}={{\tan }^{-1}}\left| \frac{-(\sqrt{3}+2)-1}{1-(\sqrt{3}+2)} \right|={{\tan }^{-1}}\left| \frac{-(\sqrt{3}+3)}{-(\sqrt{3}+1)} \right|\] \[={{\tan }^{-1}}(\sqrt{3})=60{}^\circ \]. \[{{\theta }_{23}}={{\tan }^{-1}}\left| \frac{\sqrt{3}-2-1}{1+\sqrt{3}-2} \right|={{\tan }^{-1}}\left| \frac{\sqrt{3}-3}{\sqrt{3}-1} \right|\] \[={{\tan }^{-1}}(\sqrt{3})=60{}^\circ \].
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