Question Number 14630 by Don sai last updated on 03/Jun/17 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{eqn} \\ $$$$\mathrm{dr}/\mathrm{d}\theta=\left[\mathrm{r}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \right)/\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right]\mathrm{cot}\theta \\ $$$$\mathrm{hint}.\:\mathrm{let}\:\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} . \\…
Question Number 14502 by ajfour last updated on 01/Jun/17 Commented by ajfour last updated on 01/Jun/17 $${Find}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}\:\:{in}\:{terms}\:{of}\: \\ $$$$\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}\:{using}\:{parameters} \\ $$$$\boldsymbol{{h}},\:{and}\:\boldsymbol{{k}}.\: \\ $$ Commented by…
Question Number 145573 by mnjuly1970 last updated on 06/Jul/21 Answered by Olaf_Thorendsen last updated on 06/Jul/21 $${p}\:=\:{half}\:{perimeter} \\ $$$${p}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}{r}+\mathrm{2}+\mathrm{2}{r}+\mathrm{3}+\mathrm{2}{r}\right)\:=\:\mathrm{3}\left({r}+\mathrm{1}\right) \\ $$$${p}−{a}\:=\:\mathrm{3}\left({r}+\mathrm{1}\right)−\left(\mathrm{1}+\mathrm{2}{r}\right)\:=\:{r}+\mathrm{2} \\ $$$${p}−{b}\:=\:\mathrm{3}\left({r}+\mathrm{1}\right)−\left(\mathrm{2}+\mathrm{2}{r}\right)\:=\:{r}+\mathrm{1} \\ $$$${p}−{c}\:=\:\mathrm{3}\left({r}+\mathrm{1}\right)−\left(\mathrm{3}+\mathrm{2}{r}\right)\:=\:{r}…
Question Number 145575 by mnjuly1970 last updated on 06/Jul/21 Answered by ajfour last updated on 06/Jul/21 Commented by ajfour last updated on 06/Jul/21 $${Let}\:{the}\:{intersection}\:{be}\:{the} \\…
Question Number 80015 by mr W last updated on 30/Jan/20 Commented by mr W last updated on 30/Jan/20 $$\mathrm{1}.\:{Prove}\:{that}\:{both}\:{equilaterial}\:{triangles} \\ $$$${touch}\:{each}\:{other}\:{in}\:{the}\:{square}\:{as}\:{shown}. \\ $$$$\mathrm{2}.\:{Find}\:{r}_{\mathrm{1}} /{r}_{\mathrm{2}} \\…
Question Number 14384 by ajfour last updated on 31/May/17 Commented by ajfour last updated on 31/May/17 $${only}\:{find}\:{S}={x}+{y}+{z} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:,\:\boldsymbol{{c}}\:\:{which}\:{are} \\ $$$${sides}\:{of}\:\bigtriangleup{ABC}. \\ $$ Commented by…
Question Number 14364 by RasheedSindhi last updated on 31/May/17 $$\mathrm{Modification}\:\mathrm{of}\:\mathrm{Q}#\mathrm{14157} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{xy}=\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{yz}=\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{zx}=\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{Pl}\:\mathrm{discuss}\:\mathrm{also}\:\mathrm{geometrical}/ \\…
Question Number 14365 by RasheedSindhi last updated on 31/May/17 $$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q}#\mathrm{14157} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}=\alpha^{\mathrm{2}} \\ $$$$\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{bc}=\beta^{\mathrm{2}} \\ $$$$\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} −\mathrm{cd}=\gamma^{\mathrm{2}} \\ $$$$\mathrm{d}^{\mathrm{2}} +\mathrm{e}^{\mathrm{2}}…
Question Number 145379 by ajfour last updated on 04/Jul/21 Commented by ajfour last updated on 04/Jul/21 $${The}\:{inner}\:{circles}\:{have}\:{unit}\:{radii}. \\ $$$${Find}\:{radius}\:{of}\:{outer}\:{circle}. \\ $$ Answered by mr W…
Question Number 145370 by imjagoll last updated on 04/Jul/21 $$\mathrm{There}\:\mathrm{are}\:\mathrm{two}\:\mathrm{circles}\:,\:\mathrm{C}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{and}\:\mathrm{C}_{\mathrm{r}} \: \\ $$$$\mathrm{of}\:\mathrm{radius}\:\mathrm{r}\:\mathrm{which}\:\mathrm{intersect}\:\mathrm{on}\:\mathrm{a}\:\mathrm{plain}\: \\ $$$$\mathrm{At}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{intersecting} \\ $$$$\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumferences}\:\mathrm{of} \\ $$$$\mathrm{C}\:\mathrm{and}\:\mathrm{C}_{\mathrm{r}} \:,\mathrm{the}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{C}\:\mathrm{and} \\ $$$$\mathrm{that}\:\mathrm{to}\:\mathrm{C}_{\mathrm{r}} \:\mathrm{form}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{120}°\:\mathrm{outside} \\ $$$$\mathrm{of}\:\mathrm{C}\:\mathrm{and}\:\mathrm{C}_{\mathrm{r}}…