Question

Given: △FKL, FK=a

m∠F=45°, m∠L=30°

Find: FL
I know that I need to add an altitude from K so I can get a 45, 45, 90 triangle and a 30, 60, 90 triangle, but after that, I keep getting the answer wrong.

Answer 1

Here is the portal answer. The person who did first has explanation

Answer 2

`FL= (a (√(3) + 1))/(√(2) )`

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 (a^(2) )/(2) = (a)/(√(2) )`

∴ FE = EK =
`(a)/(√(2) )`

Now in ΔEKL, EK =
`(a)/(√(2) )`

Using trigonometry ratio,

TanФ = Altitude\ Base

`Tan \theta = (KE)/(EL)`

Tan 30° =
`(a/√(2) )/(EL)`

`(1)/(√(3) ) = (a)/(√(2) EL)`

`EL = (√(3)a)/(√(2) )`

Now FL = FE + EL

FE =
`(a)/(√(2) )` and
`EL = (√(3)a)/(√(2) )`

∴
`FL = (a)/(√(2) ) + (√(3)a )/(√(2) ) = (a (√(3) + 1))/(√(2) )`