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Question Number 203151 by sonukgindia last updated on 11/Jan/24

Answered by mr W last updated on 11/Jan/24

Commented by mr W last updated on 11/Jan/24

$$\frac{AB}{\sin (\theta +60°)}=\frac{1}{\sin 60°}$$$$\Rightarrow AB=\frac{\sin \theta }{\sqrt{3}}+\cos \theta$$$$2\alpha +\theta =60°$$$$\Rightarrow \alpha =30°-\frac{\theta }{2}$$$$BD\tan 30°=DA\tan \alpha$$$$(AB-DA)\tan 30°=DA\tan (30°-\frac{\theta }{2})$$$$\Rightarrow DA=\frac{AB}{1+\sqrt{3}\tan (30°-\frac{\theta }{2})}$$$$\Rightarrow DA=\frac{\frac{\sin \theta }{\sqrt{3}}+\cos \theta }{1+\sqrt{3}\tan (30°-\frac{\theta }{2})}$$$$AE=\frac{DA}{\cos 2\alpha }=\frac{\frac{\sin \theta }{\sqrt{3}}+\cos \theta }{\cos (60°-\theta )[1+\sqrt{3}\tan (30°-\frac{\theta }{2})]}$$$$AE=\frac{2(\sin \theta +\sqrt{3}\cos \theta )}{\sqrt{3}(\cos \theta +\sqrt{3}\sin \theta )[1+\sqrt{3}\tan (30°-\frac{\theta }{2})]}$$$$AE=\frac{2(\tan \theta +\sqrt{3})}{\sqrt{3}(1+\sqrt{3}\tan \theta )[1+\sqrt{3}\tan (30°-\frac{\theta }{2})]}$$$${\left(AE\right)}_{min}\approx 0.9082\Rightarrow {\left(EC\right)}_{max}\approx 0.0918$$$$at\theta \approx 0.4317(24.73°)$$

Commented by MathematicalUser2357 last updated on 20/Jan/24

$$=\frac{2(\tan 24.73°+\sqrt{3})}{\sqrt{3}(1+\sqrt{3}\tan 24.73°)(1+\sqrt{3}\tan (30°-\frac{24.73°}{2}))}$$

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