Find unknown angles of a triangle

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Discussion Overview

The discussion revolves around finding the unknown angles of a triangle based on given conditions, including specific angle measures and the equality of certain line segments. Participants explore geometric principles and the application of the sine law to solve the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates angle ADC as 72 degrees based on the exterior angle theorem and expresses uncertainty about proceeding further.
  • Another participant suggests applying the sine law after assuming equal segments are 1 unit each, but questions whether geometry principles alone could suffice.
  • Several participants request clarification on the exact problem statement and its source, indicating potential discrepancies in the provided information.
  • Some participants assert that triangle BCD is an isosceles right triangle, claiming angles BDC and CDB are 45 degrees, while another participant disputes this characterization, stating it is not a right-angled triangle and providing a counter-argument based on arctan calculations.

Areas of Agreement / Disagreement

Participants express differing views on the nature of triangle BCD and the methods to solve for the unknown angles. There is no consensus on the correct approach or the validity of the assumptions made.

Contextual Notes

Participants note the lack of a complete problem statement and the potential for errors in the diagram or assumptions about the triangle's properties.

tantrik
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Dear friends,

I am unable to find out the unknown angles for the following triangle which I attached with this post.

View attachment 6092

Angle BAD and angle BCD are the unknown angles need to be calculated. Given that lines AB=BC=CD and angle CDE = 108 degrees

From my calculations: angle ADC = 180 - 108 = 72 degrees (angles on a straight line)

angle BAD + angle BCD = 108 degrees (exterior angle of a triangle = sum of interior opposite angles)

I could not proceed any further beyond this. I thought line DB (median) is bisecting angle ADC since line DB is bisecting line AC (AB=BC) but this only happens in case of isosceles and equilateral triangles.

I am very much stuck here and seek your kind suggestions here. I am also suspecting something could be wrong in the diagram of the triangle or may be the unknown angles. Let me know where I am wrong. Thanks in advance.
 

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Without a loss of generality, let the equal segments be equal to 1 unit each and then apply the sine law.
 
greg1313 said:
Without a loss of generality, let the equal segments be equal to 1 unit each and then apply the sine law.

Thanks but is it possible to solve this question just by using geometry principles?
 
Can you post the exact problem statement? (I'm assuming you're giving us your version of it). Where did you get the problem?
 
greg1313 said:
Can you post the exact problem statement? (I'm assuming you're giving us your version of it). Where did you get the problem?

No, this exactly how the question (diagram only, no other info about triangles or angles) has been provided by the math teacher in the school.
 
Use the fact that, since triangle BCD is an isosceles right triangle, angles BDC and CDB have measure 45 degrees.

The rest is easy.
 
HallsofIvy said:
Use the fact that, since triangle BCD is an isosceles right triangle, angles BDC and CDB have measure 45 degrees.

The rest is easy.

It's not a right-angled triangle. If it were, arctan(1/2) would be 18 degrees but arctan(1/2) is about 27 degrees.
 

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