Discern how many triangles are possible given two sides and a non-included angle

Question

For the longest time, I always thought you could only determine a triangle, having only two sides, if you also had the included angle. Recently I solved a trigonometry problem where I was given $2$ sides and a non included angle and I was able to determine the third side, which was eye opening to me. So I did some doodles to try and see why is it possible to sometimes determine a triangle given only $2$ sides and a non included angle.

What I concluded was that if you have sides $a$ and $b$ as well as the angle adjacent to $a$, you can draw side $a$ and on one end draw a circle of radius side $b$, on the other end draw a line at the given angle. What you will notice is that sometimes the circle is too small and the angle is too big so the line never touches the circle. Other times the line can touch it exactly once and other times exactly twice. Since a line cannot touch a circle more than twice, there are no more possibilities.

So I want to know, is that idea correct? Also if it is correct how do I know whether the given two sides and angle determine $0$ triangles, $1$ triangle or $2$ triangles?

I can tell that when it is $1$ triangle and side $a$ is greater than side $b$, the line drawn at an angle is tangent to the circle, hence a $90°$ degrees angle is formed, there is a right triangle where $sin(B) = \frac{b}{a}$. Where $B$ is the given angle, adjacent to side $a$ and opposite of side $b$. Of course what I am interested in is knowing whether or not the converse of that fact is true, ie if $2$ sides and a non included angle opposite of side $b$ are given and $a > b$, and $sin(B) = \frac{b}{a}$ then there exists exactly one triangle with those requirements and it's a right triangle with hypotenuse side $a$. Of course if side $a$ is equal to side $b$ there can only be at most one triangle, thinking of it as the circle engulfing side $a$. There will always be one triangle as long as the angle is lesser than $90°$. If side $a$ is less than side $b$, then side $a$ is not fully engulfed and there will always be a triangle not matter what angle (lesser than 180° of course).

So, I would like to know for all possibilities ($a > b$, $a = b$, $a < b$, etc...): How many triangles are there given sides $a$ and $b$ and angle opposite to $b$, $B$.

Answer 1

I didn't read it completely, but your intuitive idea seems reasonable. Let $a,b,c$ be the lengths of the sides of the triangle, and $\angle A,\angle B, \angle C$ be the angles opposite to them respectively.

Suppose you are given $a,b$ and $\angle B$. Draw a line segment of length $a$, and from one endpoint, draw a circle of radius $b$. Name that endpoint $C$, and the other empty endpoint $B$. We divide this into cases:

1. Case 1: $b>a$. Then only one such triangle exists (note that $B$ lies inside the circle), up to reflection.
2. Case 2: $b=a$. Then it is obvious that one such triangle exists if $\angle B <90^\circ$ and zero otherwise (as we can determine all other angles since the triangle is isosceles).
3. Case 3: $b<a$. Then if $\angle B>90^\circ$ then zero triangles. For one triangle, we must have $\sin(\angle B) = b/a$. If $B<\sin^{-1}(b/a)$, then two triangles. Also, note that if $\angle B \in (\sin^{-1}(b/a),90^\circ)$, then also zero triangles. Hence, if $\angle B>\sin^{-1}(b/a)$, then zero triangles.