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lora16 [44]
3 years ago
9

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

Mathematics
2 answers:
drek231 [11]3 years ago
8 0

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

<u>FM</u>

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

<u>ML</u>

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}  

<u>FM + ML = FL</u>

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

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

Nitella [24]3 years ago
4 0

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)}

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