COS A=\FRAC{({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2bc}\] \[\Rightarrow \cos {{60}^{o}}=\frac{1}{2}=\frac{9+{{c}^{2}}-16}{2.3c}\]Þ \[{{c}^{2}}-3c-7=0\].

"> COS A=\FRAC{({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2bc}\] \[\Rightarrow \cos {{60}^{o}}=\frac{1}{2}=\frac{9+{{c}^{2}}-16}{2.3c}\]Þ \[{{c}^{2}}-3c-7=0\].

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In a triangle\[ABC,\]\[a=4,b=3\], \[\angle A={{60}^{o}}\]. Then c is the root of the equation                       [Roorkee 1993]

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

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\[\COS A=\FRAC{({{b}^{2}}+{{c}^{2}}-{{a}^{2}})}{2bc}\] \[\Rightarrow \cos {{60}^{o}}=\frac{1}{2}=\frac{9+{{c}^{2}}-16}{2.3c}\]Þ \[{{c}^{2}}-3c-7=0\].

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Posted on 25 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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