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Question Number 156806 by MathSh last updated on 15/Oct/21

Answered by mr W last updated on 16/Oct/21

((AB)/(AC))=((sin 60)/(sin 20)) ((BC)/(AC))=((sin 80)/(sin 20)) ((DC)/(AC))=((sin 80)/(sin 30)) ((AB×DC)/(BC×AC))=((AB)/(AC))×((AC)/(BC))×((DC)/(AC)) =((sin 60)/(sin 20))×((sin 20)/(sin 80))×((sin 80)/(sin 30)) =((sin 60)/(sin 30)) =2cos 30 =(√3)

$$\frac{{AB}}{{AC}}=\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{sin}\:\mathrm{20}} \\ $$$$\frac{{BC}}{{AC}}=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{20}} \\ $$$$\frac{{DC}}{{AC}}=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{30}} \\ $$$$\frac{{AB}×{DC}}{{BC}×{AC}}=\frac{{AB}}{{AC}}×\frac{{AC}}{{BC}}×\frac{{DC}}{{AC}} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{sin}\:\mathrm{20}}×\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{80}}×\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\mathrm{30}} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{60}}{\mathrm{sin}\:\mathrm{30}} \\ $$$$=\mathrm{2cos}\:\mathrm{30} \\ $$$$=\sqrt{\mathrm{3}} \\ $$

Commented by MathSh last updated on 16/Oct/21

Very nice dear Ser thank you

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$

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