Anybody can tell me how to find y1 , y2 , angle 1

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Discussion Overview

The discussion revolves around a problem involving the determination of unknown lengths (y1, y2) and angles (angle 1, angle 2) in a geometric context, likely related to triangles. The participants explore the implications of missing information and the relationships between the variables involved.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the problem cannot be solved due to a lack of one equation, noting that there are five equations for six unknowns. They suggest that if the angle between y1 and y2 were provided, the problem could be resolved.
  • The same participant lists the equations involved, including relationships based on sine functions and the Pythagorean theorem, indicating the complexity of the problem.
  • Another participant acknowledges a mistake in their previous contributions, agreeing with the assessment of the problem's solvability.
  • A different viewpoint is presented, suggesting that Heron's formula may not add new information and questioning the necessity of a fixed vertex position, proposing that any position between the two known segments could satisfy the equations.

Areas of Agreement / Disagreement

Participants express disagreement regarding the sufficiency of the provided information to solve the problem. While some acknowledge the complexity and potential mistakes, others propose alternative interpretations of the geometric configuration.

Contextual Notes

The discussion highlights limitations related to the number of equations versus unknowns, as well as assumptions about the positioning of geometric elements that remain unresolved.

bent al physics
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Anybody can tell me how to find y1 , y2 , angle 1 and 2

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The problem cannot be solved because it lacks one equation:there can be written 5 equations for the 6 unknowns.If the angle between Y_{1} and Y_{2} were given,the problem would be solved.
Equations:
[tex]\sin\alpha_{1}=\frac{150}{Y_{1}}[/tex]
[tex]\sin\alpha_{2}=\frac{220}{Y_{2}}[/tex]
[tex]a+b=350[/tex]
[tex]a^{2}+150^{2}=Y_{1}^{2}[/tex]
[tex]b^{2}+220^{2}=Y_{2}^{2}[/tex]

As u can see,an equation is missing.There is an option of solving for 5 unknowns wrt une unknown chosen as parameter.But that wouldn't be too convenient,right??

Daniel.

PS.Alpha_{1},alpha_{2} are the two angles unknown (labeled through "angle 1" and "angle 2",respectively),while "a" and "b" are the two remaining sides of the wo trianges which have not been labeled on the picture.
 
ahrkon and daniel is absolutely right.
What a dreadful mistake on my part.
 
Last edited:
I get the impression that Heron's formula is implied by the rest (and hence brings no new information).

I don't see any reason why the vertex in the middle has to be in one position and no other. So far any position between the two segments with known size (150 and 220 m) would satisfy all equations. Isn't that so?
 

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