I know how incredibly stupid this question sounds, and I know any 2 triangles you draw with same line lengths will have the same 3 angles, but I just can't figure out the exact reason why having 3 lines of the same, or congruent, lengths, would lead to 2 triangles having the same angles. Can someone please explain?
Asked
Active
Viewed 302 times
1
-
law of cosines – achille hui Jun 18 '18 at 09:35
-
It's not stupid to question intuition. It helps you have an open mind. – qwr Jun 18 '18 at 09:44
-
My question about this, however, is does the law of cosines rely on the property above to be true? Because if it does, that doesn't prove anything. That would be circular. – Ethan Chan Jun 18 '18 at 09:47
3 Answers
2
No matter what, in order for $a$ to be the same length, point $C$ must fall on circle $d$. And no matter what, in order for $b$ to be the same length, point $C$ must fall on circle $e$.
With these restrictions, the only choices for $C$ are the two points where the circles intersect. And these choices differ only by being mirror images of each other, which doesn't change angles, only their winding.
Dan Uznanski
- 11,025
-
And distinct circles cannot intersect in more than two places as three points determine a (circum-)circle – Henry Jun 18 '18 at 11:40
1
If you are comfortable accepting the law of cosines, that provides one explanation. This theorem states that, if a triangle has sides with lengths $a, b$, and $c$ whose opposing angle measures are $\alpha, \beta$, and $\gamma$ respectively, then the following relation holds:
$$c^2 = a^2 + b^2 - 2ab \cos(\gamma)$$
Of course, we can arrive at two similar equations since there are $3$ different ways to label the sides with those letters and whichever way we choose is arbitrary. For instance, we also have:
$$b^2 = a^2 + c^2 - 2ac \cos(\beta)$$
Looking at the first, notice that we can solve for $\cos(\gamma)$ in terms of the side lengths. Because $\cos:(0,\pi) \rightarrow [-1,1]$ is an injective function, the value of $\cos(\gamma)$ determines $\gamma$ itself. To be explicit:
$$\gamma = \arccos \left( \frac{c^2 - a^2 - b^2}{-2ab} \right)$$
And we have similar equations for angles $\alpha$ and $\beta$.
Kaj Hansen
- 33,011
-
My question about this, however, is does the law of cosines rely on the property above to be true? Because if it does, that doesn't prove anything. That would be circular. – Ethan Chan Jun 18 '18 at 09:46
-
1@EthanChan, that is definitely a good thing to be concerned about. But it turns out we can prove it without needing SSS congruence. See here: https://en.wikipedia.org/wiki/Law_of_cosines#Using_the_distance_formula – Kaj Hansen Jun 18 '18 at 09:51
1
Given three lengths $a,b,c$ such that $a+b\ge c$, $a+c\ge b$ and $b+c\ge a$ we can construct the triangle uniquely up to rotation, translation and reflection: Draw points $A,B$ at distance $c$ connected by a line segment. The third point $C$ must have distance $a$ from $B$ and distance $b$ from $A$. Hence, it is an intersection point of the circle around $A$ with radius $b$ and the circle around $B$ with radius $a$. There are exactly two such points and the resulting triangles are mirror images of each other. The only choices here were where to put the line segment $AB$ and which of the two intersection points you choose. None of these choices influence the angles of the triangles.
Christoph
- 24,912