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Category: Coordinate Geometry

Question-86063

Question Number 86063 by john santu last updated on 26/Mar/20 Answered by mr W last updated on 26/Mar/20 $$\left(\frac{\mathrm{1}}{{R}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{6}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} }\right) \\…

Given-a-sphere-of-unit-radius-Find-the-expression-of-a-circular-spot-on-the-sphere-s-surface-given-the-latitude-and-the-longitude-of-its-center-and-its-angular-radius-r-

Question Number 20511 by dioph last updated on 27/Aug/17 $$\mathrm{Given}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{radius}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circular} \\ $$$$\mathrm{spot}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sphere}'\mathrm{s}\:\mathrm{surface}\:\mathrm{given} \\ $$$$\mathrm{the}\:\mathrm{latitude}\:\beta\:\mathrm{and}\:\mathrm{the}\:\mathrm{longitude}\:\lambda \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{center}\:\mathrm{and}\:\mathrm{its}\:\mathrm{angular}\:\mathrm{radius}\:{r}. \\ $$ Terms of Service Privacy Policy…

Find-two-possible-values-of-p-if-the-lines-px-y-0-and-3x-y-1-0-intersect-at-45-

Question Number 151513 by pete last updated on 21/Aug/21 $$\mathrm{Find}\:\mathrm{two}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{p}\:\mathrm{if}\:\mathrm{the}\:\mathrm{lines} \\ $$$${px}−{y}=\mathrm{0}\:\mathrm{and}\:\mathrm{3}{x}+{y}+\mathrm{1}=\mathrm{0}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{45}° \\ $$ Answered by Olaf_Thorendsen last updated on 21/Aug/21 $$\Delta_{\mathrm{1}} \::\:{px}−{y}\:=\:\mathrm{0} \\ $$$$\Delta_{\mathrm{2}}…

Question-150864

Question Number 150864 by puissant last updated on 16/Aug/21 Answered by ajfour last updated on 16/Aug/21 $$\mathrm{70}°=\theta,\:\mathrm{30}°=\alpha,\:\mathrm{40}°=\beta,\:\mathrm{6}{cm}={r} \\ $$$${A}_{\mathrm{1}} =\left({r}−\mathrm{2}{r}\mathrm{cos}\:\theta\right){r}\mathrm{sin}\:\beta \\ $$$${A}_{\mathrm{2}} =\frac{{r}\left(\mathrm{1}−\mathrm{2cos}\:\theta\right)\left\{{r}\mathrm{sin}\:\theta−{r}\mathrm{sin}\:\beta\right\}}{\mathrm{2}} \\ $$$${A}_{\mathrm{3}}…