Question Number 50829 by Raj Singh last updated on 21/Dec/18 Commented by Raj Singh last updated on 21/Dec/18 $${this}\:{is}\:{class}\:\mathrm{10}\:{problem}\:{so}\:{solve} \\ $$$${by}\:{general}\:{method} \\ $$ Answered by…
Question Number 181869 by mnjuly1970 last updated on 01/Dec/22 Commented by mr W last updated on 01/Dec/22 $${see}\:{Q}\mathrm{180582} \\ $$ Terms of Service Privacy Policy…
Question Number 181875 by mr W last updated on 01/Dec/22 $${if}\:{a}+{b}+{c}=\mathrm{0},\:{find}\:{the}\:{maximum}\:{of} \\ $$$$\frac{\mid{a}+\mathrm{2}{b}+\mathrm{3}{c}\mid}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}. \\ $$ Commented by mr W last updated on…
Question Number 181852 by mnjuly1970 last updated on 01/Dec/22 Answered by mr W last updated on 01/Dec/22 $$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}=\frac{{c}}{\mathrm{sin}\:\gamma}=\mathrm{2}{R} \\ $$$${S}_{{ABC}} =\frac{{bc}\:\mathrm{sin}\:\alpha}{\mathrm{2}}=\mathrm{2}{R}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta\:\mathrm{sin}\:\gamma \\ $$$${S}_{{ABC}} ={R}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma\right){r}…
Question Number 181839 by amin96 last updated on 01/Dec/22 Commented by mr W last updated on 01/Dec/22 $${there}\:{is}\:{no}\:{unique}\:{solution}.\:\alpha\:{can}\:{be} \\ $$$${any}\:{value}\:{between}\:\mathrm{0}\:{and}\:\mathrm{90}°. \\ $$ Commented by Acem…
Question Number 181834 by mr W last updated on 01/Dec/22 Commented by mr W last updated on 01/Dec/22 $${an}\:{unsolved}\:{old}\:{question} \\ $$ Terms of Service Privacy…
Question Number 50762 by ajfour last updated on 19/Dec/18 Commented by ajfour last updated on 19/Dec/18 $${Find}\:{radii}\:\:{r}\:{and}\:{R}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}} \\ $$$${and}\:\boldsymbol{{b}}. \\ $$ Answered by mr W…
Question Number 181812 by HeferH last updated on 01/Dec/22 Answered by mr W last updated on 01/Dec/22 $$\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{sin}\:\left(\mathrm{2}\alpha+\mathrm{30}°\right)}{\mathrm{2}\:\mathrm{cos}\:\alpha} \\ $$$$\mathrm{sin}\:\left(\mathrm{2}\alpha+\mathrm{30}°\right)=\mathrm{cos}\:\alpha=\mathrm{sin}\:\left(\mathrm{90}°−\alpha\right) \\ $$$$\mathrm{2}\alpha+\mathrm{30}°=\mathrm{90}°−\alpha\:\Rightarrow\alpha=\mathrm{20}° \\ $$$${or} \\…
Question Number 181783 by mr W last updated on 30/Nov/22 $${one}\:{side}\:{of}\:{a}\:{triangle}\:{is}\:\mathrm{20}\:{cm}.\:{the} \\ $$$${other}\:{two}\:{sides}\:{are}\:{in}\:{ratio}\:\mathrm{1}:\mathrm{3}. \\ $$$$\left.\mathrm{1}\right)\:{what}\:{is}\:{the}\:{maximum}\:{area}\:{of}\:{the} \\ $$$${triangle},\:{if}\:{exists}? \\ $$$$\left.\mathrm{2}\right)\:{what}\:{is}\:{the}\:{maximun}\:{perimeter} \\ $$$${of}\:{the}\:{triangle},\:{if}\:{exists}? \\ $$ Answered by…
Question Number 181766 by HeferH last updated on 30/Nov/22 Answered by som(math1967) last updated on 30/Nov/22 $$\:{let}\:{radius}\:{of}\:{inner}\:{circle}\:={r} \\ $$$$\:\frac{{CI}}{{r}}={cosec}\mathrm{20}\Rightarrow{CI}={rcosec}\mathrm{20} \\ $$$$\:{c}={rcot}\mathrm{30}+{rcot}\mathrm{40} \\ $$$$={r}\left(\frac{{cos}\mathrm{30}{sin}\mathrm{40}+{cos}\mathrm{40}{sin}\mathrm{30}}{{sin}\mathrm{30}{sin}\mathrm{40}}\right) \\ $$$$={r}\left(\frac{\mathrm{2}{sin}\mathrm{70}}{\mathrm{2}{sin}\mathrm{20}{cos}\mathrm{20}}\right)…