Question Number 121244 by benjo_mathlover last updated on 06/Nov/20 | ||
Commented by liberty last updated on 06/Nov/20 | ||
Commented by liberty last updated on 06/Nov/20 | ||
$$\left(\mathrm{1}\right)\:\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{2}}{\mathrm{OA}}\:\Rightarrow\:\mathrm{OA}=\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\mathrm{AD}\:=\:\sqrt{\mathrm{14}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{7}^{\mathrm{2}} \left(\mathrm{4}−\mathrm{1}\right)}=\:\mathrm{7}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{3}\right)\mathrm{OD}\:=\:\mathrm{AD}−\mathrm{OA}\:=\:\mathrm{7}\sqrt{\mathrm{3}}−\mathrm{4}\:\mathrm{cm} \\ $$$$\mathrm{7}\sqrt{\mathrm{3}}−\mathrm{4}= \\ $$$$\mathrm{8}.\mathrm{124356} \\ $$ | ||