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Question-138089

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Question Number 138089 by Bekzod Jumayev last updated on 10/Apr/21
Commented by Bekzod Jumayev last updated on 10/Apr/21
Help?
$$\boldsymbol{{Help}}? \\ $$
Answered by mr W last updated on 10/Apr/21
Commented by mr W last updated on 10/Apr/21
say AD=DC=1 ⇒AB=1 ((FA)/(sin 20°))=((FB)/(sin 60°))=((AB)/(sin 40°)) ⇒FA=((sin 20°)/(sin 40°))=(1/(2 cos 20°))=AG ⇒FB=((sin 60°)/(sin 40°))=((√3)/(2 sin 40°))=BE AG×sin 60°=(AC−AG×cos 60°)×tan ∠ACG tan ∠ACG=(((1/(2 cos 20°))×sin 60°)/(2−(1/(2 cos 20°))×cos 60°))=((√3)/(8 cos 20°−1)) AD×sin 60°=(AE−AD×cos 60°)×tan ∠AED tan ∠AED=((sin 60°)/(1+((√3)/(2 sin 40°))−cos 60°))=(((√3) sin 40°)/( (√3)+sin 40°)) ?=∠AED+∠EBH=∠AED+∠ACG+60° =60°+tan^(−1) (((√3) sin 40°)/( (√3)+sin 40°))+tan^(−1) ((√3)/(8 cos 20°−1)) =100°
$${say}\:{AD}={DC}=\mathrm{1} \\ $$$$\Rightarrow{AB}=\mathrm{1} \\ $$$$\frac{{FA}}{\mathrm{sin}\:\mathrm{20}°}=\frac{{FB}}{\mathrm{sin}\:\mathrm{60}°}=\frac{{AB}}{\mathrm{sin}\:\mathrm{40}°} \\ $$$$\Rightarrow{FA}=\frac{\mathrm{sin}\:\mathrm{20}°}{\mathrm{sin}\:\mathrm{40}°}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}°}={AG} \\ $$$$\Rightarrow{FB}=\frac{\mathrm{sin}\:\mathrm{60}°}{\mathrm{sin}\:\mathrm{40}°}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{40}°}={BE} \\ $$$${AG}×\mathrm{sin}\:\mathrm{60}°=\left({AC}−{AG}×\mathrm{cos}\:\mathrm{60}°\right)×\mathrm{tan}\:\angle{ACG} \\ $$$$\mathrm{tan}\:\angle{ACG}=\frac{\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}°}×\mathrm{sin}\:\mathrm{60}°}{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}°}×\mathrm{cos}\:\mathrm{60}°}=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}\:\mathrm{cos}\:\mathrm{20}°−\mathrm{1}} \\ $$$${AD}×\mathrm{sin}\:\mathrm{60}°=\left({AE}−{AD}×\mathrm{cos}\:\mathrm{60}°\right)×\mathrm{tan}\:\angle{AED} \\ $$$$\mathrm{tan}\:\angle{AED}=\frac{\mathrm{sin}\:\mathrm{60}°}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{40}°}−\mathrm{cos}\:\mathrm{60}°}=\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{40}°}{\:\sqrt{\mathrm{3}}+\mathrm{sin}\:\mathrm{40}°} \\ $$$$?=\angle{AED}+\angle{EBH}=\angle{AED}+\angle{ACG}+\mathrm{60}° \\ $$$$=\mathrm{60}°+\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{40}°}{\:\sqrt{\mathrm{3}}+\mathrm{sin}\:\mathrm{40}°}+\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{8}\:\mathrm{cos}\:\mathrm{20}°−\mathrm{1}} \\ $$$$=\mathrm{100}° \\ $$
Commented by Bekzod Jumayev last updated on 10/Apr/21
Thank you
$${Thank}\:{you} \\ $$
Commented by otchereabdullai@gmail.com last updated on 11/Apr/21
fantastic!
$$\mathrm{fantastic}! \\ $$

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