Find unknown angles of a triangle

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SUMMARY

The discussion focuses on calculating the unknown angles BAD and BCD in a triangle where AB=BC=CD and angle CDE is given as 108 degrees. The user correctly deduces that angle ADC equals 72 degrees and establishes that angle BAD plus angle BCD equals 108 degrees. The solution involves recognizing triangle BCD as an isosceles right triangle, leading to angles BDC and CDB each measuring 45 degrees. The user is encouraged to apply geometric principles and the sine law to further solve the problem.

PREREQUISITES
  • Understanding of triangle properties, specifically isosceles triangles
  • Knowledge of exterior angle theorem in triangles
  • Familiarity with the sine law in trigonometry
  • Basic geometry principles related to angles and lines
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  • Study the properties of isosceles triangles and their angle relationships
  • Learn about the exterior angle theorem and its applications in triangle problems
  • Explore the sine law and its use in solving triangle-related calculations
  • Practice solving geometric problems involving unknown angles and triangle properties
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Students studying geometry, mathematics educators, and anyone interested in solving triangle-related problems using geometric principles and trigonometric laws.

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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