Trig problem involving a triangle's angles and sides

  • #1
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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)
 
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Answers and Replies

  • #2
verty
Homework Helper
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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.
 
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Likes Kolika28
  • #3
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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 :)
 
  • #4
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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?
 
  • #5
SammyS
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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) \,.##
 
  • #6
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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 :)
 

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