COS \theta =\frac{4+6-{{(\sqrt{3}+1)}^{2}}}{2.2.\sqrt{6}}\RIGHTARROW \theta ={{75}^{o}}\].

"> COS \theta =\frac{4+6-{{(\sqrt{3}+1)}^{2}}}{2.2.\sqrt{6}}\RIGHTARROW \theta ={{75}^{o}}\].

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If the sides of a triangle are in the ratio \[2:\sqrt{6}:(\sqrt{3}+1)\], then the largest angle of the triangle will be [MP PET 1990]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 11 months ago

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\[\COS \theta =\frac{4+6-{{(\sqrt{3}+1)}^{2}}}{2.2.\sqrt{6}}\RIGHTARROW \theta ={{75}^{o}}\].

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