Question Number 65312 by Tony Lin last updated on 28/Jul/19 Commented by Tony Lin last updated on 28/Jul/19 $${how}\:{to}\:{know}\:{x}=\mathrm{5} \\ $$ Commented by mr W…
Question Number 65281 by mr W last updated on 27/Jul/19 Commented by mr W last updated on 27/Jul/19 $${segment}\:{of}\:{circle}\:{with}\:{size}\:\mathrm{2}{c}×{d}.\:\left({d}\leqslant{c}\right) \\ $$$${find}\:{the}\:{inscribed}\:{ellipse}\:{with} \\ $$$${maximum}\:{area}. \\ $$…
Question Number 65235 by ajfour last updated on 26/Jul/19 Commented by ajfour last updated on 26/Jul/19 $${Find}\:{p},{q},{r}\:{in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$ Commented by ajfour last updated on…
Question Number 65212 by ajfour last updated on 26/Jul/19 Commented by ajfour last updated on 26/Jul/19 $${For}\:{P}\:{to}\:{be}\:{unique},\:{determine} \\ $$$${how}\:{a},{b},{c}\:{are}\:{related}. \\ $$ Answered by ajfour last…
Question Number 65189 by arcana last updated on 26/Jul/19 $${all}\:{square}\:{is}\:{a}\:{rhombus}.\:{why}? \\ $$ Answered by MJS last updated on 26/Jul/19 $$\mathrm{a}\:\mathrm{rhombus}\:\mathrm{has}\:\mathrm{2}\:\mathrm{parallel}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{sides},\:\mathrm{all} \\ $$$$\mathrm{of}\:\mathrm{them}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{length} \\ $$$$\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{a}\:\mathrm{special}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{rhombus} \\…
Question Number 65153 by Tawa1 last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19 Commented by Tawa1 last updated on 25/Jul/19 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}…
Question Number 65148 by Tawa1 last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19…
Question Number 65128 by Tawa1 last updated on 25/Jul/19 Answered by ajfour last updated on 25/Jul/19 Commented by ajfour last updated on 25/Jul/19 $${OP}\:=\mathrm{8} \\…
Question Number 65119 by Tawa1 last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19…
Question Number 64971 by Tawa1 last updated on 23/Jul/19 Commented by Tony Lin last updated on 23/Jul/19 $$\frac{\frac{{x}}{\mathrm{2}}}{{sin}\theta}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{sin}\left(\mathrm{90}°+\theta\right)}=\frac{{x}}{{sin}\left(\mathrm{90}°−\mathrm{2}\theta\right)} \\ $$$$\Rightarrow\frac{\frac{{x}}{\mathrm{2}}}{{sin}\theta}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{cos}\theta}=\frac{{x}}{{cos}\mathrm{2}\theta} \\ $$$${let}\:{cos}\theta={t} \\ $$$$\Rightarrow\frac{\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{t}}=\frac{{x}}{\mathrm{2}{t}^{\mathrm{2}}…