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Question Number 14863 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented by adelson last updated on 05/Jun/17

$${what}^{\prime }s\mathit{W}?$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$$ABCD,square.$$$$1)showthat:{DE}^2+{EB}^2={AE}^2+{EC}^2.$$$$2)if:AE=EB=AB\Rightarrow \measuredangle CED=?$$

Commented by RasheedSoomro last updated on 05/Jun/17

$$\mathrm{AE}=\mathrm{EB}=\mathrm{AB}\Rightarrow \mathrm{CE}=\mathrm{DE}$$$$\mathrm{The}\mathrm{diagram}\mathrm{is}\mathrm{not}\mathrm{exactly}\mathrm{according}\mathrm{to}$$$$\mathrm{data}.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$that:ifAE=EB=AB.$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$figureofmrAjfouriscorrect.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$$if:t=z=a\Rightarrow h_2=a\frac{\sqrt{3}}{2},h_4=a-a\frac{\sqrt{3}}{2}$$$$tg\measuredangle CDE=\frac{h_4}{\frac{a}{2}}=\frac{a-a\frac{\sqrt{3}}{2}}{\frac{a}{2}}=2-\sqrt{3.}$$$$\Rightarrow \measuredangle CDE=\measuredangle EDC={15}^{\bullet }$$$$\Rightarrow \measuredangle DEC=180-2\times 15={150}^{\bullet }.◼$$$$$$

Answered by ajfour last updated on 05/Jun/17

$${DE}^2+{EB}^2=x^2+y^2+{(a-x)}^2+{(a-y)}^2$$$${AE}^2+{EC}^2={(a-x)}^2+y^2+x^2+{(a-y)}^2$$$$thereforeequal.$$$$$$$$When△AEBisequilateral,$$$$a-y=yory=\frac{a}{2}$$$$a-x=a\cos \frac{\pi }{6}=\frac{a\sqrt{3}}{2}$$$$\Rightarrow x=a(1-\frac{\sqrt{3}}{2})=a\left(\frac{2-\sqrt{3}}{2}\right)$$$$\mathrm{\angle }CED={2\tan }^{-1}\left(\frac{a-y}{x}\right)$$$$={2\tan }^{-1}\left(\frac{y}{x}\right)$$$$={2\tan }^{-1}\left[\frac{\left(a/2\right)}{a(2-\sqrt{3})/2)}\right]$$$$={2\tan }^{-1}(2+\sqrt{3})$$$$or\mathrm{\angle }CED=\pi -{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$$$$=\pi -\frac{\pi }{6}=\frac{5\pi }{6}.$$

Commented by ajfour last updated on 05/Jun/17

Commented by RasheedSoomro last updated on 05/Jun/17

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{quicker}!$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$$thankyoumrAjfour!youransweris$$$$quickandniceandcorrectandsmart!$$$$godblessyoumyfriend.$$

Commented by ajfour last updated on 05/Jun/17

$$TanksofthanksSir.$$

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