Question Number 139645 by otchereabdullai@gmail.com last updated on 30/Apr/21 | ||
Commented by mr W last updated on 30/Apr/21 | ||
$${OD}={AO}×\mathrm{sin}\:\angle{OAB}=\mathrm{52}×\frac{\mathrm{5}}{\mathrm{13}}=\mathrm{20} \\ $$$${BD}=\sqrt{\mathrm{25}^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} }=\mathrm{15} \\ $$$${BC}=\mathrm{2}×{BD}=\mathrm{2}×\mathrm{15}=\mathrm{30} \\ $$ | ||
Commented by otchereabdullai@gmail.com last updated on 30/Apr/21 | ||
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{profW} \\ $$ | ||