Question-14863

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

{w h a t}^{'} s W ?

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

A B C D, s q u a r e .

1) s h o w t h a t : {D E}^{2} + {E B}^{2} = {A E}^{2} + {E C}^{2} .

2) i f : A E = E B = A B \Rightarrow ∡ C E D = ?

Commented by RasheedSoomro last updated on 05/Jun/17

AE = EB = AB \Rightarrow CE = DE

The diagram is not exactly according to

data .

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

You can't use 'macro parameter character #' in math mode

t h a t : i f A E = E B = A B .

You can't use 'macro parameter character #' in math mode

f i g u r e o f m r A j f o u r i s c o r r e c t .

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

i f : t = z = a \Rightarrow h_{2} = a \frac{\sqrt{3}}{2}, h_{4} = a - a \frac{\sqrt{3}}{2}

t g ∡ C D E = \frac{h_{4}}{\frac{a}{2}} = \frac{a - a \frac{\sqrt{3}}{2}}{\frac{a}{2}} = 2 - \sqrt{3 .}

\Rightarrow ∡ C D E = ∡ E D C = 15^{∙}

\Rightarrow ∡ D E C = 180 - 2 \times 15 = 150^{∙} .

Answered by ajfour last updated on 05/Jun/17

{D E}^{2} + {E B}^{2} = x^{2} + y^{2} + {(a - x)}^{2} + {(a - y)}^{2}

{A E}^{2} + {E C}^{2} = {(a - x)}^{2} + y^{2} + x^{2} + {(a - y)}^{2}

t h e r e f o r e e q u a l .

W h e n △ A E B i s e q u i l a t e r a l,

a - y = y o r y = \frac{a}{2}

a - x = a \cos \frac{π}{6} = \frac{a \sqrt{3}}{2}

\Rightarrow x = a (1 - \frac{\sqrt{3}}{2}) = a (\frac{2 - \sqrt{3}}{2})

∠ C E D = {2 \tan}^{- 1} (\frac{a - y}{x})

= {2 \tan}^{- 1} (\frac{y}{x})

= {2 \tan}^{- 1} [\frac{(a / 2)}{a (2 - \sqrt{3}) / 2)}]

= {2 \tan}^{- 1} (2 + \sqrt{3})

o r ∠ C E D = π - \tan^{- 1} (\frac{1}{\sqrt{3}})

= π - \frac{π}{6} = \frac{5 π}{6} .

Commented by ajfour last updated on 05/Jun/17

Commented by RasheedSoomro last updated on 05/Jun/17

{You}^{'} re quicker!

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

t h a n k y o u m r A j f o u r! y o u r a n s w e r i s

q u i c k a n d n i c e a n d c o r r e c t a n d s m a r t!

g o d b l e s s y o u m y f r i e n d .

Commented by ajfour last updated on 05/Jun/17

T a n k s o f t h a n k s S i r .

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