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Question-139645




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

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