TAN \left( \frac{B-C}{2} \right)=\frac{b-c}{b+c}\COT \frac{A}{2}\] Þ \[\tan \left( \frac{90{}^\circ }{2} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\cot \frac{A}{2}\] Þ \[\tan \left( \frac{A}{2} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}\] Þ \[\frac{A}{2}={{15}^{o}}\Rightarrow A={{30}^{o}}\].

"> TAN \left( \frac{B-C}{2} \right)=\frac{b-c}{b+c}\COT \frac{A}{2}\] Þ \[\tan \left( \frac{90{}^\circ }{2} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\cot \frac{A}{2}\] Þ \[\tan \left( \frac{A}{2} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}\] Þ \[\frac{A}{2}={{15}^{o}}\Rightarrow A={{30}^{o}}\].

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If in a triangle \[ABC\], \[b=\sqrt{3}\], \[c=1\] and \[B-C={{90}^{o}}\]then \[\angle A\] is [MP PET 1983]

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

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\[\TAN \left( \frac{B-C}{2} \right)=\frac{b-c}{b+c}\COT \frac{A}{2}\] Þ \[\tan \left( \frac{90{}^\circ }{2} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\cot \frac{A}{2}\] Þ \[\tan \left( \frac{A}{2} \right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}\] Þ \[\frac{A}{2}={{15}^{o}}\Rightarrow A={{30}^{o}}\].

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Posted on 12 Sep 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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