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

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22-09-2018
Mathematics

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

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tramserran tramserran · Answer 1 · 2018-09-22T09:46:04+02:00

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 = [tex]\frac{a}{\sqrt{2}} = \frac{a\sqrt{2} }{2}[/tex]

ML

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

FM + ML = FL

[tex](\frac{3}{3})\frac{a\sqrt{2} }{2} + \frac{a\sqrt{6} }{6}[/tex]

= [tex]\frac{3a\sqrt{2} + a\sqrt{6} }{6}[/tex]

aachen aachen · Answer 2 · 2019-09-21T20:39:14+02:00

Answer:

[tex]\sqrt{2}\frac{a(\sqrt{3}+2)}{(\sqrt{3}+1)}[/tex]

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[tex]\text{FLP}[/tex],

[tex]\text{KP}=\text{x}[/tex]

[tex]\text{FP}=\text{a}+\text{x}[/tex]

as[tex]\text{m}\angle \text{PFL}=45^{\circ}=\text{m}\angle\text{FLP}[/tex]

[tex]\text{FP}=\text{a}+\text{x}=\text{LP}[/tex]

now, in Δ[tex]\text{KLP}[/tex],

[tex]\text{m}\angle\text{PKL}=75^{\circ}[/tex]

[tex]\text{tan}75=\frac{\text{LP}}{\text{PK}}[/tex]

[tex]\text{KP}=\text{x}[/tex]

[tex]\text{LP}=\text{x}+\text{a}[/tex]

[tex]2+\sqrt{3}=\frac{\text{a}+\text{x}}{\text{x}}[/tex]

[tex]2\text{x}+\text{x}\sqrt{3}=\text{a}+\text{x}[/tex]

[tex]\text{x}=\frac{\text{a}}{(\sqrt{3}+1)}[/tex]

[tex]\text{LP}=\frac{\text{a}}{(\sqrt{3}+1)}+\text{a}[/tex]

[tex]\text{LP}=\frac{\text{a}(\sqrt{3}+2)}{(\sqrt{3}+1)}[/tex]

now, in Δ[tex]\text{FLP}[/tex]

[tex]\text{LP}=\text{FP}[/tex]

using pythagoras theorem

[tex]\text{LP}^2+\text{FP}^2=\text{FL}^2[/tex]

[tex]2\text{LP}^2=\text{FL}^2[/tex]

[tex]\text{FL}^2=2\frac{a^2(\sqrt{3}+2)^2}{(\sqrt{3}+1)^2}[/tex]

[tex]\text{FL}=\sqrt{2}\frac{a(\sqrt{3}+2)}{(\sqrt{3}+1)}[/tex]

So, length of [tex]\text{FL}[/tex] is [tex]\sqrt{2}\frac{a(\sqrt{3}+2)}{(\sqrt{3}+1)}[/tex]