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Question Number 68305 by TawaTawa last updated on 08/Sep/19 Commented by ajfour last updated on 10/Sep/19 Commented by ajfour last updated on 10/Sep/19 $${Area}\:\:{S}_{{req}} =\underset{\smile}…
Question Number 68289 by Rasheed.Sindhi last updated on 08/Sep/19 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{parts} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{By} \\ $$$$\:\mathrm{An}\:\mathrm{arc}\:\mathrm{with}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathcal{D}{etermine} \\ $$$$\:\:\left({a}\right)\:{The}\:{length}\:{of}\:{the}\:{arc} \\ $$$$\:\:\left({b}\right){The}\:{ratio}\:{in}\:{which}\:{the}\:{arc} \\ $$$$\:\:\:\:\:\:\:\:{divides}\:{the}\:{diameter}\: \\ $$$$\:\:\:\:\:\:\:\:{meeting}\:{the}\:{center}\:{of}\:{the}\:{arc}. \\…
Question Number 68183 by ajfour last updated on 06/Sep/19 Commented by ajfour last updated on 06/Sep/19 $${QBC}\:{and}\:{BPD}\:{are}\:{tangent}\:{and} \\ $$$${normal}\:{respectively}\:{to} \\ $$$${the}\:{cubic}\:{function}\:{y}={x}^{\mathrm{3}} −\mathrm{19}{x}+\mathrm{30} \\ $$$${at}\:{one}\:{of}\:{its}\:{root}\:{and}\:{QAD}\:{and} \\…
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Question Number 133679 by Ar Brandon last updated on 23/Feb/21 Commented by Ar Brandon last updated on 23/Feb/21 $$\mathrm{If}\:\mathrm{M}\:\mathrm{is}\:\mathrm{orthocenter},\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{Area}\left[\mathrm{ABC}\right]}{\mathrm{Area}\left[\mathrm{A}''\mathrm{B}''\mathrm{C}''\right]}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by…
Question Number 68132 by mr W last updated on 05/Sep/19 Commented by kaivan.ahmadi last updated on 05/Sep/19 $${a}+\beta+\theta=\mathrm{180}\:\:\:\left(\mathrm{1}\right) \\ $$$$\alpha+\beta−{a}=\mathrm{0}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}\alpha−\mathrm{2}\theta+{b}=\mathrm{0}\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{3}\right)−\left(\mathrm{2}\right):\:\alpha−\mathrm{2}\theta−\beta+\left({a}+{b}\right)=\mathrm{0}\Rightarrow\mathrm{2}\theta+\beta−\alpha=\mathrm{120}\:\:\:\left(\mathrm{4}\right) \\…
Question Number 68110 by TawaTawa last updated on 05/Sep/19 Commented by kaivan.ahmadi last updated on 07/Sep/19 $$\left(\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow{y}=\pm\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$${equation}\:{of}\:{BC}: \\ $$$${m}={tg}\mathrm{30}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 68092 by TawaTawa last updated on 04/Sep/19 Commented by TawaTawa last updated on 05/Sep/19 $$\mathrm{I}\:\mathrm{understand}\:\mathrm{very}\:\mathrm{well}\:\mathrm{now}\:\mathrm{sir}. \\ $$ Commented by kaivan.ahmadi last updated on…
Question Number 68041 by A8;15: last updated on 03/Sep/19 Commented by mr W last updated on 03/Sep/19 $${see}\:{Q}#\mathrm{67386} \\ $$ Terms of Service Privacy Policy…