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Question-136760




Question Number 136760 by I want to learn more last updated on 25/Mar/21
Answered by mr W last updated on 25/Mar/21
((BC)/(sin ∠BDC))=((CD)/(sin 30°))     ...(i)  ((sin ∠BDA)/(AB))=((sin 45°)/(DA))   ...(ii)  (i)×(ii):  ((BC)/(AB))=((sin 45°)/(sin 30°))=(√2)
$$\frac{{BC}}{\mathrm{sin}\:\angle{BDC}}=\frac{{CD}}{\mathrm{sin}\:\mathrm{30}°}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\angle{BDA}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{45}°}{{DA}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{{BC}}{{AB}}=\frac{\mathrm{sin}\:\mathrm{45}°}{\mathrm{sin}\:\mathrm{30}°}=\sqrt{\mathrm{2}} \\ $$
Commented by I want to learn more last updated on 25/Mar/21
I appreciate sir. God bless you
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Mar/21
thanks
$$\mathrm{thanks} \\ $$

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