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Question Number 97005 by PRITHWISH SEN 2 last updated on 06/Jun/20

Commented by malwaan last updated on 06/Jun/20

$$\mathit{a}=102-\measuredangle \mathrm{ACB}$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{I}\mathrm{think}\mathrm{some}\mathrm{information}\mathrm{is}\mathrm{missing}.\mathrm{I}\mathrm{want}\mathrm{your}$$$$\mathrm{sugestions}.$$

Answered by mr W last updated on 06/Jun/20

Commented by mr W last updated on 06/Jun/20

$$sayAD=1$$$$AB=BD=\frac{1}{2\cos 72}$$$$\frac{AD}{\sin 66}=\frac{CD}{\sin 30}$$$$\Rightarrow CD=\frac{AD\sin 30}{\sin 66}=\frac{1}{2\sin 66}$$$$BC=\sqrt{{CD}^2+{BD}^2-2\times CD\times BD\times \cos 12}$$$$BC=\sqrt{\frac{1}{4{\sin }^{2}66}+\frac{1}{4{\cos }^{2}72}-\frac{\cos 12}{2\sin 66\cos 72}}$$$$\frac{\sin \alpha }{CD}=\frac{\sin 12}{BC}$$$$\sin \alpha =\frac{CD\sin 12}{BC}$$$$=\frac{\sin 12}{2\sin 66\sqrt{\frac{1}{4{\sin }^{2}66}+\frac{1}{4{\cos }^{2}72}-\frac{\cos 12}{2\sin 66\cos 72}}}$$$$=\frac{\sin 12}{\sqrt{1+\frac{{\sin }^{2}66}{{\cos }^{2}72}-\frac{2\sin 66\cos 12}{\cos 72}}}$$$$\alpha ={\sin }^{-1}\frac{\sin 12}{\sqrt{1+\frac{{\sin }^{2}66}{{\cos }^{2}72}-\frac{2\sin 66\cos 12}{\cos 72}}}$$$$=6°$$

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{superb}!\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}.$$

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