Menu Close

Category: Coordinate Geometry

Question-150864

Tinku Tara 0 Comments

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}}…

Read more →

R-r-2-R-2-R-r-2-R-2-r-2-2rR-R-2-R-2-r-2-2rR-R-2-4rR-r-R-4-1cm-A-S-pir-2-pi-1cm-2-picm-2-

Tinku Tara 0 Comments

Question Number 150793 by cherokeesay last updated on 15/Aug/21 $$\left({R}−{r}\right)^{\mathrm{2}} \:+\:{R}^{\mathrm{2}} \:=\:\left({R}\:+\:{r}\right)^{\mathrm{2}} \:\Leftrightarrow \\ $$$${R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} −\mathrm{2}{rR}\:+\:{R}^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} \:+\:{r}^{\mathrm{2}} \:+\mathrm{2}{rR}\:\Rightarrow \\ $$$${R}^{\mathrm{2}} \:=\:\mathrm{4}{rR}\:\Rightarrow\:{r}\:=\:\frac{{R}}{\mathrm{4}}\:=\:\mathrm{1}{cm} \\ $$$$\mathscr{A}_{{S}}…

Read more →