Question #bf8db

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Question #bf8db

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Answer 1 · 2016-02-03T23:30:37

${63.435}^{\circ}$ $63.435^@$

Explanation:

Consider the figure below

I created this figure using MS Excel

We know that
$3 V E = E K$ $3VE=EK$ or $E K = 3 V E$ $EK=3VE$
And
$V E + E K = V K$ $VE+EK=VK$
Then

$V E + 3 V E = V K$ $VE+3VE=VK$ => $V E = \frac{V K}{4}$ $VE=(VK)/4$
Since $V K = \sqrt{2} \cdot s$ $VK=sqrt(2)*s$ => $V E = \frac{\sqrt{2}}{4} \cdot s$ $VE=sqrt(2)/4*s$

Since $△_{A F V} \equiv △_{F I V} \equiv △_{F I K} \equiv △_{A F K}$ $triangle_(AFV)-=triangle_(FIV)-=triangle_(FIK)-=triangle_(AFK)$
$\to F V = \frac{V K}{2}$ $-> FV=(VK)/2$ => $F V = \frac{\sqrt{2}}{2} \cdot s$ $FV=sqrt(2)/2*s$ (and $F I = F V$ $FI=FV$ )
$\to E F = F V - V E = \frac{\sqrt{2}}{2} \cdot s - \frac{\sqrt{2}}{4} \cdot s$ $-> EF=FV-VE=sqrt(2)/2*s-sqrt(2)/4*s$ => $E F = \frac{\sqrt{2}}{4} \cdot s$ $EF=sqrt(2)/4*s$

In $△_{E F I}$ $triangle_(EFI)$

$E I^{2} = E F^{2} + F I^{2} = {(\frac{\sqrt{2}}{4} \cdot s)}^{2} + {(\frac{\sqrt{2}}{2} \cdot s)}^{2} = (\frac{2}{16} + \frac{2}{4}) s^{2} = \frac{2 + 8}{16} \cdot s^{2}$ $EI^2=EF^2+FI^2=(sqrt(2)/4*s)^2+(sqrt(2)/2*s)^2=(2/16+2/4)s^2=(2+8)/16*s^2$ => $E I = \frac{\sqrt{10}}{4} \cdot s$ $EI=sqrt(10)/4*s$

Still in $△_{E F I}$ $triangle_(EFI)$ , using Law of Sines:

$\frac{E I}{{sin 90}^{\circ}} = \frac{F I}{sin F ˆ E I}$ $(EI)/(sin 90^@)=(FI)/(sin F hat E I)$ => $sin F hat E I=(FI*1)/(EI)=(sqrt(2)/2*cancel(s))/(sqrt(10)/4*cancel(s))=2*sqrt(2/10)=2/sqrt(5)$
=> $F hat E I=sin^(-1) (2/sqrt(5))=63.435^@$

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Question #bf8db

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