SINCE the angles are in A.P., THEREFORE \[B={{60}^{o}}\] and \[\frac{b}{c}=\frac{\sin B}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}}{2\sin C}=\frac{\sqrt{3}}{\sqrt{2}}\] Þ \[C={{45}^{o}}\]so that \[A={{180}^{o}}-{{60}^{o}}-{{45}^{o}}={{75}^{o}}\]. "> SINCE the angles are in A.P., THEREFORE \[B={{60}^{o}}\] and \[\frac{b}{c}=\frac{\sin B}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}}{2\sin C}=\frac{\sqrt{3}}{\sqrt{2}}\] Þ \[C={{45}^{o}}\]so that \[A={{180}^{o}}-{{60}^{o}}-{{45}^{o}}={{75}^{o}}\]. ">
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If in a triangle the angles are in A. P. and \[b:c=\sqrt{3}:\sqrt{2}\], then \[\angle A\]is equal to  [IIT 1981; Kurukshetra CEE 1998; Pb. CET 1990]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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SINCE the angles are in A.P., THEREFORE \[B={{60}^{o}}\] and \[\frac{b}{c}=\frac{\sin B}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{3}}{2\sin C}=\frac{\sqrt{3}}{\sqrt{2}}\] Þ \[C={{45}^{o}}\]so that \[A={{180}^{o}}-{{60}^{o}}-{{45}^{o}}={{75}^{o}}\].
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Posted on 27 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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