Finding Angle ACB in Triangle ABC

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Discussion Overview

The discussion revolves around finding the angle ACB in triangle ABC, given specific angle measures and a point D on line segment BC. The problem involves geometric reasoning and potentially different approaches to arrive at the solution.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Post 1 presents the problem statement with given angles and relationships between segments.
  • Post 2 proposes a method involving a point M between A and D, leading to the conclusion that angle ACB is 75 degrees.
  • Post 3 reiterates the approach from Post 2, confirming the conclusion of angle ACB being 75 degrees.
  • Post 5 asserts that angle ACB can be calculated as the sum of 30 degrees and 45 degrees, also arriving at 75 degrees.
  • Post 6 references a diagram to support the claim that angle ACB equals 75 degrees, echoing the previous points.

Areas of Agreement / Disagreement

Participants generally agree that angle ACB is 75 degrees based on different reasoning approaches, but the discussion does not explore alternative methods or challenge the conclusions presented.

Contextual Notes

The discussion relies on geometric relationships and assumptions about the configuration of triangle ABC and point D. The reasoning presented does not explore potential variations or alternative configurations that could affect the outcome.

Who May Find This Useful

Readers interested in geometric problem-solving, particularly in the context of triangle properties and angle calculations, may find this discussion beneficial.

Albert1
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$\triangle ABC , \angle ABC=45^o,\,\, point \,\, D\,\, on \,\, \,\overline{BC} $

$and,\,\, 2\overline{BD}=\overline{CD},\,\, \angle DAB=15^o$

$find :\,\, \angle ACB=?$
 
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Albert said:
$\triangle ABC , \angle ABC=45^o,\,\, point \,\, D\,\, on \,\, \,\overline{BC} $

$and,\,\, 2\overline{BD}=\overline{CD},\,\, \angle DAB=15^o$

$find :\,\, \angle ACB=?$
Find a point $M$ between $A$ and $D$ such that $\angle MBD=30$. Note that $MD=DM$ and $AM=MB$. Say $BD=1$. The above leads to $AD=\sqrt 3 + 1$. Further note that $\angle CMD=90$ and hence $\angle MCA=45$. Consequently $\angle ACB=75$.
 
caffeinemachine said:
Find a point $M$ between $A$ and $D$ such that $\angle MBD=30$. Note that $MD=DM$ and $AM=MB$. Say $BD=1$. The above leads to $AD=\sqrt 3 + 1$. Further note that $\angle CMD=90$ and hence $\angle MCA=45$. Consequently $\angle ACB=75$.
caffeinemachine :very good solution :cool:
 
Albert said:
caffeinemachine :very good solution :cool:
Thanks. :) If you have a different one then please post it.
 

Attachments

  • Angle ACB.jpg
    Angle ACB.jpg
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Albert said:
https://www.physicsforums.com/attachments/1209
from the diagram it is easy to see that :
$\angle ACB =30^o +45^o =75^o$
Awesome!
 

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