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Question Number 37989 by ajfour last updated on 20/Jun/18

Commented by ajfour last updated on 20/Jun/18

$$\left(i\right)Findareaofbluesquareand$$$$redsquare.$$$$\left(ii\right)Find\theta forwhichcornersof$$$$redsquareliesonsidesofblue$$$$square.$$

Commented by MrW3 last updated on 20/Jun/18

$$\left(i\right)$$$$sidelengthofbluesquare:$$$$a_1=\frac{a}{\cos \theta }-a\sin \theta (1+\tan \theta )$$$$\Rightarrow a_1=a(\cos \theta -\sin \theta )$$$$\Rightarrow A_1=a_1^2=a^2(1-\sin 2\theta )$$$$$$$$sidelengthofredsquare:$$$$a_2=a\tan \theta \sin \alpha with$$$$\tan \alpha =\frac{a}{a-a\tan \theta }=\frac{\cos \theta }{\cos \theta -\sin \theta }$$$$or\sin \alpha =\frac{\cos \theta }{\sqrt{1+{\cos }^2\theta -\sin 2\theta }}$$$$\Rightarrow a_2=a\frac{\sin \theta }{\sqrt{1+{\cos }^2\theta -\sin 2\theta }}$$$$\Rightarrow A_2=a_2^2=a^2\frac{{\sin }^2\theta }{1+{\cos }^2\theta -\sin 2\theta }$$$$$$$$\left(ii\right)$$$$a_1=a_2[\sin (\alpha -\theta )+\cos (\alpha -\theta )]$$$$a_1=a_2[\sin \alpha (\sin \theta +\cos \theta )+\cos \alpha (\cos \theta -\sin \theta )]$$$$(\cos \theta -\sin \theta )=\frac{\sin \theta }{\sqrt{1+{\cos }^2\theta -\sin 2\theta }}[\frac{\cos \theta (\sin \theta +\cos \theta )}{\sqrt{1+{\cos }^2\theta -\sin 2\theta }}+\frac{{(\cos \theta -\sin \theta )}^2}{\sqrt{1+{\cos }^2\theta -\sin 2\theta }}]$$$$\cos \theta -\sin \theta =\frac{\sin \theta (\sin \theta \cos \theta +{\cos }^2\theta +1-\sin 2\theta )}{1+{\cos }^2\theta -\sin 2\theta }$$$$\cos \theta -\sin \theta =\frac{{\sin }^2\theta \cos \theta }{1+{\cos }^2\theta -\sin 2\theta }+\sin \theta$$$$2(\cos \theta -2\sin \theta )=\frac{\sin \theta \sin 2\theta }{1+{\cos }^2\theta -\sin 2\theta }$$$$2(\cos \theta -2\sin \theta )(1+{\cos }^2\theta +\sin 2\theta )=\sin \theta \sin 2\theta$$$$$$$$\Rightarrow \theta \approx 23.3°$$

Commented by ajfour last updated on 20/Jun/18

$$Thankyousir,pleaseanswer\left(ii\right)$$$$even..$$

Commented by MrW3 last updated on 20/Jun/18

$$Icansolve\left(ii\right)onlynumerically.$$

Commented by ajfour last updated on 20/Jun/18

$$yesSir,thisisright.ThankYou.$$

Commented by MrW3 last updated on 20/Jun/18

$$seecorrection.\theta \approx 23.3°.$$

Commented by MrW3 last updated on 20/Jun/18

Answered by ajfour last updated on 20/Jun/18

$$letsideofbluesquarebe\mathit{b},and$$$$thatofredsquarebe\mathit{r}.$$$$\mathit{b}=\mathit{a}(\cos \mathit{\theta }-\sin \mathit{\theta })$$$$m=\frac{a}{a-a\tan \theta }=\frac{1}{1-\tan \mathit{\theta }}$$$$\mathit{r}=\frac{\mathit{a}\tan \mathit{\theta }}{\sqrt{1+{(1-\tan \mathit{\theta })}^2}}.$$

Answered by ajfour last updated on 20/Jun/18

$$letbottomcornerofredsquare$$$$beB(h,k)andm=\frac{a}{a-a\tan \theta }$$$$\Rightarrow 1+m\tan \theta =m$$$$\frac{k}{h-a\tan \theta }=m\left(i\right)$$$$\frac{k}{a-h}=\frac{1}{m}....\left(ii\right)$$$$ifBliesonlowersideofblue$$$$square,thenk=h\tan \theta ...\left(iii\right)$$$$from\left(ii\right)and\left(iii\right):$$$$mh\tan \theta =a-h$$$$\Rightarrow h=\frac{a}{1+m\tan \theta };k=\frac{a\tan \theta }{1+m\tan \theta }$$$$since1+m\tan \theta =m$$$$\mathit{h}=\frac{\mathit{a}}{\mathit{m}};\mathit{k}=\frac{\mathit{a}\tan \mathit{\theta }}{\mathit{m}}$$$$Usingin\left(i\right):$$$$\frac{a\tan \theta }{m}=a-am\tan \theta$$$$\tan \theta =m-m^2\tan \theta$$$$\tan \theta =\frac{1-\frac{\tan \theta }{(1-\tan \theta )}}{1-\tan \theta }$$$$let\tan \theta =s$$$$\Rightarrow s{(1-s)}^2=1-2s$$$$\theta ={\tan }^{-1}s.$$$$(Ingoodagreementwithyour$$$$answer,Sir).$$

Commented by MrW3 last updated on 20/Jun/18

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