Question

Given: △FKL, FK=a, m∠F=45°, m∠L=30° Find: FL

Answer 1

The best way to do this is to draw a picture of ΔFKL and include line segment KM that is perpendicular to FL.  This creates ΔFKM which is a 45°-45°-90° triangle and ΔLKM which is a 30°-60°-90° triangle.

Find the lengths of FM and ML.  Then, FM + ML = FL

FM

ΔFKM (45°-45°-90°): FK is the hypotenuse so FM = $\frac{a}{\sqrt{2}} = \frac{a\sqrt{2} }{2}$

ML

ΔLKM (30°-60°-90°): from ΔFKM, we know that KM = $\frac{a\sqrt{2} }{2}$ , so KL = $\frac{a}{\sqrt{6}} = \frac{a\sqrt{6} }{6}$  

FM + ML = FL

$(\frac{3}{3})\frac{a\sqrt{2} }{2} + \frac{a\sqrt{6} }{6}$

= $\frac{3a\sqrt{2} + a\sqrt{6} }{6}$

Answer 2

Answer:

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

Step-by-step explanation:

Given: △FKL, FK=a, m∠F=45°, m∠L=30°

To Find:

Solution: Consider the file attached.

 a perpendicular is drawn to Line FK from L, LP

 in Δ$\text{FLP}$,

$\text{KP}=\text{x}$

$\text{FP}=\text{a}+\text{x}$  

as$\text{m}\angle \text{PFL}=45^{\circ}=\text{m}\angle\text{FLP}$

$\text{FP}=\text{a}+\text{x}=\text{LP}$

now, in Δ$\text{KLP}$,

$\text{m}\angle\text{PKL}=75^{\circ}$

$\text{tan}75=\frac{\text{LP}}{\text{PK}}$

$\text{KP}=\text{x}$

$\text{LP}=\text{x}+\text{a}$

$2+\sqrt{3}=\frac{\text{a}+\text{x}}{\text{x}}$

$2\text{x}+\text{x}\sqrt{3}=\text{a}+\text{x}$

$\text{x}=\frac{\text{a}}{(\sqrt{3}+1)}$

$\text{LP}=\frac{\text{a}}{(\sqrt{3}+1)}+\text{a}$

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

now, in Δ$\text{FLP}$

$\text{LP}=\text{FP}$

using pythagoras theorem

$\text{LP}^2+\text{FP}^2=\text{FL}^2$

$2\text{LP}^2=\text{FL}^2$

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

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

So, length of $\text{FL}$ is $\sqrt{2}\frac{a(\sqrt{3}+2)}{(\sqrt{3}+1)}$