Animated Solution for Mathematics - Trigonometry: There exists a triangle $ABC$ satisfying the conditions

Welcome, fellow explorer of the mathematical universe! Today, we are not just solving a problem; we are embarking on a journey to understand the very conditions that allow a triangle to exist.

Imagine you are standing on a vast, flat plane. You have a fixed angle $A$ and a fixed side length $b$. Your goal is to place a third vertex $B$ such that a triangle $ABC$ is formed with a side $a$ opposite to angle $A$.

But wait—can you always form this triangle? That is the question we are going to answer.

Our First Guardian

To begin, let's fix vertex $C$ and angle $A$. We drop a perpendicular from $C$ to the base line where $B$ must lie. This perpendicular is our altitude, $h$.

Using basic trigonometry, we find that:

This value, $h=b\sin A$, is the threshold of existence. It is the shortest distance from $C$ to the base line. If side $a$ is shorter than this, it simply cannot reach the base. It will hang in the air, a ghost of a triangle that never was.

The Algebraic Bridge

Now, how do we connect this geometric intuition to the algebraic conditions? We turn to the Sine Rule:

Rearranging this, we get:

Notice the beauty here: the numerator is exactly our altitude $h$. So, $\sin B=\frac{h}{a}$. This equation is the heart of our problem. It tells us that the existence of angle $B$—and thus the triangle—depends entirely on the ratio of the altitude $h$ to the side $a$.

We can now categorize all possible scenarios into three distinct worlds:

1. The Impossible World ($h>a$): If $b\sin A>a$, then $\sin B=\frac{h}{a}>1$. Since the sine of an angle cannot exceed $1$, no such angle $B$ exists. The side $a$ is too short.

2. The Right-Angled World ($h=a$): If $b\sin A=a$, then $\sin B=1$, which means $B=\frac{\pi }{2}$. We have a perfect right-angled triangle.

For this to be valid, the sum of angles $A+B$ must be less than $\pi$. Since $B=\frac{\pi }{2}$, we must have $A<\frac{\pi }{2}$. This gives us our first valid condition: $b\sin A=a$ with $A<\frac{\pi }{2}$.

3. The Ambiguous World ($h<a$): If $b\sin A<a$, then $\sin B<1$. This is where things get exciting!

If $b>a$, the side $a$ is long enough to reach the base at two different points, creating two possible triangles. This is the famous ambiguous case. For these triangles to exist on the right side of $A$, $A$ must be acute ($A<\frac{\pi }{2}$).

By carefully analyzing the relationship between the altitude $h$ and the side $a$, we have navigated through the constraints of triangle existence. We have seen how the Sine Rule acts as a bridge between geometry and algebra.

Keep this visualization in mind—the altitude as a threshold—and you will never be trapped by these questions again. Happy calculating!