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Question Number 188219 by Rupesh123 last updated on 26/Feb/23

Answered by mr W last updated on 26/Feb/23

Commented by mr W last updated on 27/Feb/23

$$2\beta +\alpha =\frac{\pi }{2}\Rightarrow \beta =\frac{\pi }{4}-\frac{\alpha }{2}$$$$lengthofrectangle=a$$$$heightofrectangle=b$$$$a=2+4\cos (30°-\alpha )$$$$b=2+4\sin (30°-\alpha )$$$$2(\sqrt{3}+\frac{1}{\tan \beta })\sin \alpha =1+\frac{1}{\tan \beta }$$$$2(\sqrt{3}+\frac{1+\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}})\times \frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}=1+\frac{1+\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}}$$$$2(\sqrt{3}+\frac{1+t}{1-t})\times \frac{2t}{1-t^2}=1+\frac{1+t}{1-t}$$$$(2\sqrt{3}-3)t^2-2(1+\sqrt{3})t+1=0$$$$t=\tan \frac{\alpha }{2}=\frac{1+\sqrt{3}-\sqrt{7}}{2\sqrt{3}-3}$$$$\Rightarrow \alpha =2{\tan }^{-1}\left(\frac{1+\sqrt{3}-\sqrt{7}}{2\sqrt{3}-3}\right)\approx 21.067630$$$$a\approx 5.951489$$$$b\approx 2.621074$$$$A\approx 15.599293$$

Commented by Rupesh123 last updated on 26/Feb/23

Very detailed solution! Perfect Explanation!

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