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Question-195799

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Question Number 195799 by sonukgindia last updated on 10/Aug/23 Answered by som(math1967) last updated on 11/Aug/23 $$\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}}{{a}}}\:+\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}}}=\mathrm{2}{a} \\ $$$$\Rightarrow\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}{a} \\ $$$$\Rightarrow{a}\left[\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\right]=\mathrm{2}{a}…

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Question-195742

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Question Number 195742 by sonukgindia last updated on 09/Aug/23 Answered by MM42 last updated on 09/Aug/23 $$\angle\alpha_{{i}} =\mathrm{120}\Rightarrow\angle{BAC}=\angle{BDF}=\mathrm{90} \\ $$$${AB}={AC}\Rightarrow\angle{C}=\angle{B}_{\mathrm{1},\mathrm{2}} =\mathrm{45}\Rightarrow\angle{B}_{\mathrm{2}} =\angle{B}_{\mathrm{3}} =\mathrm{15} \\ $$$$\frac{{x}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{90}}\:\:\:\&\:\:\:\frac{{y}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{60}}…

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