Question Number 63663 by Tawa1 last updated on 07/Jul/19 Answered by mr W last updated on 07/Jul/19 $$\angle{ADB}=\mathrm{360}−\left(\mathrm{180}−\mathrm{6}−\mathrm{30}\right)−\left(\mathrm{180}−\mathrm{6}−\mathrm{24}\right)=\mathrm{66}° \\ $$$$\frac{{AD}}{{DC}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{6}°} \\ $$$$\frac{{BD}}{{DC}}=\frac{\mathrm{sin}\:\mathrm{24}°}{\mathrm{sin}\:\mathrm{6}°} \\ $$$$\Rightarrow\frac{{AD}}{{BD}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{24}°} \\…
Question Number 129174 by Study last updated on 13/Jan/21 Commented by Study last updated on 13/Jan/21 $${x}=?\:\:\:\:\:\:\:\left[{y}=???\right. \\ $$ Answered by mr W last updated…
Question Number 129105 by I want to learn more last updated on 12/Jan/21 $$\int_{\:\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\sqrt{\mathrm{tan}\:\mathrm{x}\left(\mathrm{1}\:\:−\:\:\mathrm{tan}\:\mathrm{x}\right)}\:\:\mathrm{dx} \\ $$ Answered by Lordose last updated on 13/Jan/21…
Question Number 63564 by Tawa1 last updated on 05/Jul/19 Answered by MJS last updated on 05/Jul/19 $$\mathrm{the}\:\mathrm{flower}\:\mathrm{has}\:\mathrm{got}\:\mathrm{12}\:\mathrm{arcs},\:\mathrm{each}\:\mathrm{one}\:\mathrm{of}\:\mathrm{length} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{perimeter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\Rightarrow\: \\ $$$${p}=\mathrm{12}×\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{2}\pi{r}=\mathrm{4}\pi{r} \\ $$$${r}=\mathrm{10}\:\Rightarrow\:{p}=\mathrm{40}\pi \\ $$…
Question Number 129096 by Gulnoza last updated on 12/Jan/21 Answered by talminator2856791 last updated on 15/Jan/21 $$\:\mathrm{2}\left(\frac{\mathrm{300}−\mathrm{75}\pi}{\mathrm{4}}\right) \\ $$ Commented by Gulnoza last updated on…
Question Number 129077 by Study last updated on 12/Jan/21 Commented by benjo_mathlover last updated on 12/Jan/21 $$\frac{\mathrm{FG}}{\mathrm{8}}\:=\:\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow\mathrm{FG}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$\frac{\mathrm{DE}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{6}}\Rightarrow\mathrm{DE}=\mathrm{4} \\ $$$$\mathrm{Area}\:\mathrm{FGED}\:=\:\frac{\mathrm{8}×\mathrm{6}}{\mathrm{2}}−\frac{\frac{\mathrm{20}}{\mathrm{3}}+\mathrm{8}}{\mathrm{2}}×\mathrm{1}−\frac{\mathrm{3}×\mathrm{4}}{\mathrm{2}}\: \\ $$$$\:=\:\mathrm{24}−\left(\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{4}\right)−\mathrm{6} \\ $$$$\:=\:\mathrm{18}−\frac{\mathrm{22}}{\mathrm{3}}\:=\:\frac{\mathrm{32}}{\mathrm{3}}…
Question Number 63438 by ajfour last updated on 04/Jul/19 Commented by ajfour last updated on 04/Jul/19 $${Determine}\:\:{a}:{b}:{c}\:\:{if}\:{the} \\ $$$${inscribed}\:\bigtriangleup{ABC}\:{has}\:{constant} \\ $$$${perimeter}\:{but}\:{maximum}\:{area}. \\ $$ Commented by…
Question Number 128947 by ajfour last updated on 11/Jan/21 Commented by ajfour last updated on 11/Jan/21 $${Find}\:{radius}\:{of}\:{the}\:{blue}\:{arc}. \\ $$ Answered by mr W last updated…
Question Number 128944 by ajfour last updated on 11/Jan/21 Commented by ajfour last updated on 11/Jan/21 $${Find}\:{radius}\:{ratio}\:\frac{{R}}{{r}}. \\ $$$${The}\:{polygon}\:{is}\:{a}\:{regular} \\ $$$${pentagon}. \\ $$ Answered by…
Question Number 63324 by pradyot_pathak last updated on 02/Jul/19 Answered by Hope last updated on 02/Jul/19 $${area}\:{of}\:{pentagon}=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{5}×\mathrm{12}^{\mathrm{2}} ×{cot}\left(\frac{\mathrm{180}^{{o}} }{\mathrm{5}}\right) \\ $$$$=\mathrm{180}×{cot}\mathrm{36}^{{o}} \\ $$$${external}\:{angle}=\frac{\mathrm{360}^{{o}} }{\mathrm{5}}=\mathrm{72}^{{o}} \\…