Question

Question #bf8db

Answer 1

`63.435^@`

Explanation:

Consider the figure below

We know that
`3VE=EK` or `EK=3VE`
And
`VE+EK=VK`
Then

`VE+3VE=VK` => `VE=(VK)/4`
Since `VK=sqrt(2)*s` => `VE=sqrt(2)/4*s`

Since `triangle_(AFV)-=triangle_(FIV)-=triangle_(FIK)-=triangle_(AFK)`
`-> FV=(VK)/2` => `FV=sqrt(2)/2*s` (and `FI=FV`)
`-> EF=FV-VE=sqrt(2)/2*s-sqrt(2)/4*s` => `EF=sqrt(2)/4*s`

In `triangle_(EFI)`

`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` => `EI=sqrt(10)/4*s`

Still in `triangle_(EFI)`, using Law of Sines:

`(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^@`