Trig problem involving a triangle's angles and sides

[Kolika28]

Homework Statement

In the triangle ABC, angle A = 30 deg, angle B = 45 deg and since AC = a. The perpendicular from C on AB cuts AB in E. In this task you should calculate exact values.

a) Determine AE, BC and AB expressed by a.

b) Perpendicular from B on AC intersects the extension of AC in D. Find CD expressed by a.

c) A square KLMN is inscribed in the triangle ABC. The corners K and L are on the side AB, the corner M on BC and the corner N on AC. Find the sides in the square expressed by a.

Homework Equations

I can't find a solution to c) Can someone help me?

The Attempt at a Solution

a) AE=(a√3)2 BC=(a√2)/2 AB=(a/2)(√3+1)
b) CD=(a/4)(√3-1)

 

[verty]

I know of a way to solve this.

1. Express the length AB in terms of a.
2. Call length KN x. Express the lengths AK and LB in terms of x.

 

[Kolika28]

1. Isn't that equal to the equation I wrote in the solution over?
2. I'm not sure how I am supposed to express the lengths in terms of x. Should I use cosine?

Cos(30)=AK/AN? But I don't know what AN is equal to. I really appreciate your help by the way :)

 

[Kolika28]

I think I have an idea, but I'm not sure if this i correct.

tan(30)=KN/AK 1/√3=x/AK AK=√3*x

tan(45)=LM/LB LB=x

AB=AK+KL+LB
(a/2)(√3+1)=√3*x+x+x

I then find out that x=-(a-√3*a)/2=(a(√3-1)/2). Is this right?

 

[SammyS]

I think I have an idea, but I'm not sure if this i correct.

tan(30)=KN/AK 1/√3=x/AK AK=√3*x

tan(45)=LM/LB LB=x

AB=AK+KL+LB
(a/2)(√3+1)=√3*x+x+x

I then find out that x=-(a-√3*a)/2=(a(√3-1)/2). Is this right?

Yes. I also get that that the length of each side of the square is $\ (a/2)(\sqrt 3 - 1) \,.$

 

[Kolika28]

Yes. I also get that that the length of each side of the square is $\ (a/2)(\sqrt 3 - 1) \,.$

Thank you so much for the help :)