OBVIOUS. TRICK: Obviously it is not an EQUILATERAL triangle because A=B=C=60o does not SATISFY the given condition. But B = 90o then \[{{\sin }^{2}}B=1\] and \[{{\cos }^{2}}A+{{\cos }^{2}}C={{\cos }^{2}}A+{{\cos }^{2}}\left( \frac{\pi }{2}-A \right)\,\]                                         \[=\,{{\cos }^{2}}A+{{\sin }^{2}}A=1\] Hence this satisfy the condition, so it is a right ANGLE triangle but not necessarily isosceles.

"> OBVIOUS. TRICK: Obviously it is not an EQUILATERAL triangle because A=B=C=60o does not SATISFY the given condition. But B = 90o then \[{{\sin }^{2}}B=1\] and \[{{\cos }^{2}}A+{{\cos }^{2}}C={{\cos }^{2}}A+{{\cos }^{2}}\left( \frac{\pi }{2}-A \right)\,\]                                         \[=\,{{\cos }^{2}}A+{{\sin }^{2}}A=1\] Hence this satisfy the condition, so it is a right ANGLE triangle but not necessarily isosceles.

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If \[{{\cos }^{2}}A+{{\cos }^{2}}C={{\sin }^{2}}B,\]then \[\Delta ABC\]is [MP PET 1991]

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

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It is OBVIOUS. TRICK: Obviously it is not an EQUILATERAL triangle because A=B=C=60o does not SATISFY the given condition. But B = 90o then \[{{\sin }^{2}}B=1\] and \[{{\cos }^{2}}A+{{\cos }^{2}}C={{\cos }^{2}}A+{{\cos }^{2}}\left( \frac{\pi }{2}-A \right)\,\]                                         \[=\,{{\cos }^{2}}A+{{\sin }^{2}}A=1\] Hence this satisfy the condition, so it is a right ANGLE triangle but not necessarily isosceles.

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Posted on 31 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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