# Trig problem involving a triangle's angles and sides

• Kolika28
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)

Last edited by a moderator:

Homework Helper
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
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?

Staff Emeritus
Homework Helper
Gold Member
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 :)