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Question Number 177205 by HeferH last updated on 02/Oct/22

Answered by mr W last updated on 02/Oct/22

∠CAD=180−120−3x=60−3x ∠CDB=180−3x applying law of sines: ((sin (60−3x))/(CD))=((sin 120)/(AD)) ...(i) ((CD)/(sin 2x))=((CB)/(sin (180−3x))) ...(ii) AD=CB (i)×(ii): ((sin (60−3x))/(sin 2x))=((sin 120)/(sin (180−3x))) ((sin (60−3x))/(sin 2x))=((√3)/(2 sin 3x)) ⇒x≈8.02°

$$\angle{CAD}=\mathrm{180}−\mathrm{120}−\mathrm{3}{x}=\mathrm{60}−\mathrm{3}{x} \\ $$$$\angle{CDB}=\mathrm{180}−\mathrm{3}{x} \\ $$$${applying}\:{law}\:{of}\:{sines}: \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{60}−\mathrm{3}{x}\right)}{{CD}}=\frac{\mathrm{sin}\:\mathrm{120}}{{AD}}\:\:\:...\left({i}\right) \\ $$$$\frac{{CD}}{\mathrm{sin}\:\mathrm{2}{x}}=\frac{{CB}}{\mathrm{sin}\:\left(\mathrm{180}−\mathrm{3}{x}\right)}\:\:\:...\left({ii}\right) \\ $$$${AD}={CB} \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{60}−\mathrm{3}{x}\right)}{\mathrm{sin}\:\mathrm{2}{x}}=\frac{\mathrm{sin}\:\mathrm{120}}{\mathrm{sin}\:\left(\mathrm{180}−\mathrm{3}{x}\right)} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{60}−\mathrm{3}{x}\right)}{\mathrm{sin}\:\mathrm{2}{x}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{3}{x}} \\ $$$$\Rightarrow{x}\approx\mathrm{8}.\mathrm{02}° \\ $$

Commented by Tawa11 last updated on 02/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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