Question

Given: △FKL, FK=a

Answer 1

Answer:

$FL= \frac{a (\sqrt{3} + 1)}{\sqrt{2} }$

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,

$(FK)^{2} = (FE)^{2} + (KE)^{2}$  

$(a)^{2} = (x)^{2} + (x)^{2}$

$(a)^{2} = 2x^{2}$

$x = \sqrt \frac{a^{2} }{2} = \frac{a}{\sqrt{2} }$

∴ FE = EK = $\frac{a}{\sqrt{2} }$

Now in ΔEKL, EK = $\frac{a}{\sqrt{2} }$

Using trigonometry ratio,

TanФ = Altitude\ Base

$Tan \theta = \frac{KE}{EL}$

Tan 30° = $\frac{a/\sqrt{2} }{EL}$

$\frac{1}{\sqrt{3} } = \frac{a}{\sqrt{2} EL}$

$EL = \frac{\sqrt{3}a}{\sqrt{2} }$

Now FL = FE + EL

FE = $\frac{a}{\sqrt{2} }$ and $EL = \frac{\sqrt{3}a}{\sqrt{2} }$

∴ $FL = \frac{a}{\sqrt{2} } + \frac{\sqrt{3}a }{\sqrt{2} } = \frac{a (\sqrt{3} + 1)}{\sqrt{2} }$