In the diagram above, square $ABCD$ has a side length of $1$ and the inscribed $\triangle{FBE}$
has $m\angle{FBE} = 45^{\circ}$ and the $m\angle{BEC} = 70^{\circ}$.
Find the perimeter of $\triangle{FED}$.
+130081 angle CBE = 90 -70 = 20°
tan 20 = CE / 1 = CE
So ED = 1 - CE = 1 - tan 20
And angle FBA = 90 - 20 - 45 = 25°
tan 25 = AF/ 1 = AF
So FD = 1 - AF = 1 - tan 25
And....by the Pythagorean Theorem
FE = sqrt [ ( 1 - tan 20)^2 + ( 1 - tan 25)^2 ]
So....the perimeter of FED =
ED + FD + FE = (1 - tan 20) +( 1 - tan 25) + sqrt [ ( 1 - tan 20)^2 + ( 1 - tan 25)^2 ] =
2 units !!!
