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 =
`(a)/(√(2)) = (a√(2) )/(2)`

ML

ΔLKM (30°-60°-90°): from ΔFKM, we know that KM =
`(a√(2) )/(2)` , so KL =
`(a)/(√(6)) = (a√(6) )/(6)`

FM + ML = FL

`((3)/(3))(a√(2)  )/(2) + (a√(6)  )/(6)`

=
`(3a√(2) + a√(6) )/(6)`

Answer 2

`√(2)(a(√(3)+2))/((√(3)+1))`

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+√(3)=\frac{\text{a}+\text{x}}{\text{x}}`

`2\text{x}+\text{x}√(3)=\text{a}+\text{x}`

`\text{x}=\frac{\text{a}}{(√(3)+1)}`

`\text{LP}=\frac{\text{a}}{(√(3)+1)}+\text{a}`

`\text{LP}=\frac{\text{a}(√(3)+2)}{(√(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(a^2(√(3)+2)^2)/((√(3)+1)^2)`

`\text{FL}=√(2)(a(√(3)+2))/((√(3)+1))`

So, length of
`\text{FL}` is
`√(2)(a(√(3)+2))/((√(3)+1))`