Answer:
[tex]FL= \frac{a (\sqrt{3} + 1)}{\sqrt{2} }[/tex]
Step-by-step explanation:
As given in figure 1 below:
FK = a, m∠F = 45° and m∠L = 30°
Construction: Draw an altitude KE from point K on FL.
Now, In ΔFEK,
FE = EK (Sides opposite to equal angles of a triangle)
Let FE = EK = x
Now, using pythagoras In ΔFEK,
[tex](FK)^{2} = (FE)^{2} + (KE)^{2}[/tex]
[tex](a)^{2} = (x)^{2} + (x)^{2}[/tex]
[tex](a)^{2} = 2x^{2}[/tex]
[tex]x = \sqrt \frac{a^{2} }{2} = \frac{a}{\sqrt{2} }[/tex]
∴ FE = EK = [tex]\frac{a}{\sqrt{2} }[/tex]
Now in ΔEKL, EK = [tex]\frac{a}{\sqrt{2} }[/tex]
Using trigonometry ratio,
TanФ = Altitude\ Base
[tex]Tan \theta = \frac{KE}{EL}[/tex]
Tan 30° = [tex]\frac{a/\sqrt{2} }{EL}[/tex]
[tex]\frac{1}{\sqrt{3} } = \frac{a}{\sqrt{2} EL}[/tex]
[tex]EL = \frac{\sqrt{3}a}{\sqrt{2} }[/tex]
Now FL = FE + EL
FE = [tex]\frac{a}{\sqrt{2} }[/tex] and [tex]EL = \frac{\sqrt{3}a}{\sqrt{2} }[/tex]
∴ [tex]FL = \frac{a}{\sqrt{2} } + \frac{\sqrt{3}a }{\sqrt{2} } = \frac{a (\sqrt{3} + 1)}{\sqrt{2} }[/tex]
