Question Number 193037 by DAVONG last updated on 02/Jun/23
Answered by Subhi last updated on 02/Jun/23
$${DC}^{\mathrm{2}} ={DB}.{AD} \\ $$$$\mathrm{20}^{\mathrm{2}} ={DB}.\left(\mathrm{9}+{DB}\right) \\ $$$${DB}^{\mathrm{2}} +\mathrm{9}{DB}−\mathrm{400}=\mathrm{0} \\ $$$${DB}=\mathrm{16} \\ $$$$\frac{\mathrm{16}}{{sin}\left({DCB}\right)}=\frac{\mathrm{20}}{{sin}\left({DBC}\right)}\:\:\:\left({sin}\:{law}\right) \\ $$$$\frac{{sin}\left({DCB}\right)}{{sin}\left({DBC}\right)}=\frac{\mathrm{16}}{\mathrm{20}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${D}\hat {{C}B}={B}\hat {{A}C}\:\:\left({The}\:{same}\:{arc}\right) \\ $$$${sin}\left({D}\hat {{C}B}\right)={sin}\left({B}\hat {{A}C}\right) \\ $$$$\mathrm{180}−{D}\hat {{B}C}={A}\hat {{B}C} \\ $$$${sin}\left(\mathrm{180}−{D}\hat {{B}C}\right)={sin}\left({D}\hat {{B}C}\right)={sin}\left({A}\hat {{B}C}\right) \\ $$$$\frac{{sin}\left({DCB}\right)}{{sin}\left({DBC}\right)}=\frac{{sin}\left({BAC}\right)}{{sin}\left({ABC}\right)}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\frac{{BC}}{{sin}\left({BAC}\right)}=\frac{{AC}}{{sin}\left({ABC}\right)} \\ $$$$\frac{{BC}}{{AC}}=\frac{{sin}\left({BAC}\right)}{{sin}\left({ABC}\right)}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$ \\ $$
Commented by York12 last updated on 03/Jun/23
$${sir}\:{where}\:{are}\:{you}\:{from} \\ $$
Commented by Subhi last updated on 03/Jun/23
$${Egypt}\:,{sir} \\ $$
Answered by HeferH last updated on 02/Jun/23
$${say}:\:{BC}={a},\:{AC}={b},\:{BD}={x} \\ $$$$\mathrm{20}^{\mathrm{2}} ={x}\left(\mathrm{9}+{x}\right) \\ $$$$\:{x}^{\mathrm{2}} +\mathrm{25}{x}−\mathrm{16}{x}−\mathrm{400}=\mathrm{0} \\ $$$$\:\left({x}−\mathrm{16}\right)\left({x}+\mathrm{25}\right)=\mathrm{0} \\ $$$$\:{x}=\mathrm{16} \\ $$$$\:\frac{{a}}{{b}}\:=\:\frac{\mathrm{20}}{\mathrm{9}+{x}}\:=\:\frac{\mathrm{20}}{\mathrm{25}}\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$