Can You Find All Triangle Angles from 2 Sides?

  • Thread starter Thread starter skrat
  • Start date Start date
  • Tags Tags
    Angles Triangle
Click For Summary
SUMMARY

Determining all angles in a triangle based solely on the lengths of two sides is not possible without additional information. The discussion highlights the use of the Law of Cosines, represented as c^2 = a^2 + b^2 - 2ab \cos \gamma, and the Law of Sines, expressed as \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}, as tools for solving triangle problems. However, without knowing at least one angle, there are infinitely many configurations for the triangle, leading to multiple solutions. The conclusion is that additional parameters are necessary to uniquely determine all triangle angles.

PREREQUISITES
  • Understanding of the Law of Cosines
  • Familiarity with the Law of Sines
  • Basic knowledge of trigonometric functions
  • Concept of triangle properties and definitions
NEXT STEPS
  • Study the Law of Cosines in-depth to understand its applications in triangle problems
  • Explore the Law of Sines and its derivations for solving triangles
  • Learn about the concept of triangle congruence and similarity
  • Investigate geometric constructions involving triangles using compass and straightedge
USEFUL FOR

Students studying geometry, mathematics educators, and anyone interested in solving triangle-related problems in trigonometry.

skrat
Messages
740
Reaction score
8

Homework Statement


Is it possible to determine all the angles in a triangle, if we only know the length of two sides?

Homework Equations


The Attempt at a Solution


I was thinking for quite some time and I don't think it is possible. It probably is, if two sides are peprendicular but if not, I don't think so.
 
Last edited:
Physics news on Phys.org
Think about how you relate the sides and angles of a triangle?
 
You mean the definition of the dot product?

##\vec{a}\cdot \vec{b}=\left \| \vec{a} \right \|\left \| \vec{b} \right \|cos\theta ##

Would be great yeah, but I don't have the coordinates. I only have the length of the sides.
 
A modified version of that dot product called the cosine rule comes in handy for this. Have you studied this?
 
I have.

##c^2=a^2+b^2-2abcos\theta ##

But this is a "system" of one equations with two parameters. How would you reduce the number of parameters, or better; how would you find the length of the third side?
 
Don't you know all the sides, as in the problem?
 
Hah. Ok, there is a mistake in the original post. I apologize.
I only know the length of TWO sides. (I will edit my first post)
 
You can use the law of Sines as well. I think you can eliminate the third side by using an expression for it derived from law of sines.
 
I can eliminate the third side but than I get another angle inside the equation.

##\frac{a}{sin\alpha }=\frac{b}{sin\beta }=\frac{c}{sin\gamma }##
and
##c^2=a^2+b^2-2abcos\gamma##

gives me ##(a\frac{sin\gamma }{sin\alpha })^2=a^2+b^2-2abcos\gamma##
 
  • #10
Let ##b, c## be two sides of a triangle with known lengths and let ##\alpha## be the angle between them. Now consider each ##\alpha \in (0, \pi)##.
 
  • #11
Not at all possile to know the angle of triangle with two sides known.There will e infinite number of solutions .
Just think how will you first draw the trianle with two lengths are known.First draw one line whose length is known.Then try to draw the second line starting from on edge of the first line.This second line can be drawn at any angle zero to 360 deg.So that will result in infinite number of lines .So finally finished triange will have will have infinite solutions.
 
  • #12
If you know the length of two sides and the angle between those two, you can figure it out. If you know the length of two sides and the angle between one of them and the third side, you can narrow it down to two possibilities. If you don't know /any/ angles, though, there's nothing you can do.
 
  • #13
Yup, I thought this may be the case yeah. :/

Ok, thanks!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K