Can a triangle have an angle of 0 degrees?
This is part of a series on List of Common Misconceptions.
True or False?
An angle in a triangle can measure \(0 ^ \circ\).
Why some people say it's true: Sure, although a \(0 ^ \circ\) angle would make for a very flat triangle, or it could be very thin and pointy, if the top angle is the \(0 ^ \circ\) one.
Why some people say it's false: When we try to draw such a triangle, we only get a line which cannot be a triangle!
The statement is \( \color{red}{\textbf{false}}\).
Proof :
The key reason that triangles cannot have angles which measure \(0 ^ \circ\) is that such a figure would necessarily be 3 points on a straight line, and such a figure is not a triangle. Recall that the triangle inequality states that in a triangle, we have \( a + b > c \).
We will prove that the statement "an angle in a triangle can measure \(0 ^ \circ\)" is false using proof by contradiction. Assume that such a triangle exists, with side lengths \(a, b, \) and \(c \). Let \(c\) be the side opposite to the zero angle, then by the cosine rule
\[\begin{array}ac^2&=&a^2&+&b^2-&2ab\cos(0)\\&=&a^2&+&b^2-&2ab(1)\\&=&(a&-&b)^2.\end{array}\]
Since the length of the sides of a triangle are non-negative, we take the principle square root of both sides to get
\[c=a-b\Longrightarrow a=b+c.\]
This contradicts the triangle inequality, and hence such a triangle does not exist.
Rebuttal: These flat triangles that you're talking about are "trivial cases" of triangles, but they are still, by definition, triangles.
Reply: While it's not too hard to extend many common definitions of triangles to include objects such as the one above and on the right, all consistent definitions of triangles omit such objects because, in many ways, they don't behave as triangles do. For example, the non-triangle above is not even a polygon because it fails the part of the definition which requires that, for all line segments that define the edges of any polygon, "pairs of segments only touch at their endpoints."
Rebuttal: Consider a right triangle and the "SOH CAH TOA" identity which states that, for any right triangle with a non-right angle \(x\) adjacent to the hypotenuse, the measure of \(\sin(x) \) is the ratio of the hypotenuse of the right triangle to the side opposite the angle: \(\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{O}{H}\). Therefore, since the \(\sin\) function can take on the value of \(0\), when it does, this right-angled triangle will have one angle equal to \(0^\circ\).
Reply: If \(\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = 0,\) then it must be the case that \(O\), the side opposite the angle \(x,\) has measure 0. Therefore, the object being described is not a triangle.
Rebuttal: Since we use the ratio of the sides of a right triangle with hypotenuse 1 to define the \(\sin\) function, we need to consider "trivial triangles" which are basically just flat lines in order to make the evaluation of \(\sin(0^\circ)=0\) make any sense.
Reply: The definition of the trigonometric function being the ratio of certain sides of a right angle \(\Big(\)SOH \(\Rightarrow \sin = \frac{\text{opposite}}{\text{hypotenuse}}\Big)\) is only valid for \(0<\text{argument}<\pi\). For arguments not in that range, trigonometric functions are calculated by the unit circle.
True or False?
A triangle can have zero area.
What is the largest measure that an internal angle in a triangle might have?
See Also