Animated Solution for Mathematics - Trigonometry: In a triangle $ABC$, $AD$ is the altitude from $A$. Given $b>c$, $\angle C=23^{\circ }$ and $AD=\frac{abc}{b^2-c^2}$, then $\angle B=..................$ degrees.

Welcome, fellow traveler of the JEE Advanced journey. Today, we are not just solving a problem; we are uncovering a hidden truth about triangles.

You have been given a triangle $ABC$ with an altitude $AD$ defined by a rather intimidating expression:

At first glance, this looks like a mess of variables. But in the world of competitive mathematics, complexity is often just a mask for elegance. Let us peel back that mask.

First, let us ground ourselves. We are given $\angle C=23^{\circ }$ and $b>c$.

If you were to draw a standard acute triangle, the altitude $AD$ would fall neatly inside. But look at the denominator $b^2-c^2$. If $b>c$, this is positive.

However, the very nature of this relationship suggests something unusual. As we proceed, you will realize that $\angle B$ is obtuse. This means our altitude $AD$ does not fall inside the triangle; it falls on the extension of the line segment $BC$.

We need to find a way to connect the given formula to the properties of the triangle. We know from basic trigonometry in the right-angled triangle $△ADC$ that $AD=b\sin C$.

We now have two expressions for the same length $AD$:

Equating these two, we get $b\sin C=\frac{abc}{b^2-c^2}$. Since $b$ is a side length, it cannot be zero. We can safely divide both sides by $b$, leaving us with a much cleaner relationship:

This is our first major victory. We have stripped away the altitude and are now looking at a relationship between the sides and the angle $C$.

Now, we need to bring in the heavy artillery. The Sine Law is the bridge between sides and angles. We know that $\frac{a}{\sin A}=\frac{c}{\sin C}$.

Rearranging this, we get $\frac{\sin C}{c}=\frac{\sin A}{a}$. Let us manipulate our previous equation $\sin C=\frac{ac}{b^2-c^2}$ by dividing both sides by $c$:

Look at the left side! It is exactly what the Sine Law gives us. Therefore, we can equate the right sides:

Cross-multiplying gives us the beautiful result: $a^2=(b^2-c^2)\sin A$. We are no longer dealing with altitudes; we are dealing with the fundamental structure of the triangle.

To solve this, we must convert the side lengths $a,b,c$ into trigonometric terms. Using the circumradius $R$, we know $a=2R\sin A$, $b=2R\sin B$, and $c=2R\sin C$.

Substituting these into our equation $a^2=(b^2-c^2)\sin A$, we get:

Expanding this, the $4R^2$ terms cancel out beautifully, leaving us with:

Since $\sin A eq 0$, we divide by $\sin A$ to get $\sin A={\sin }^{2}B-{\sin }^{2}C$. Now, recall the identity ${\sin }^{2}B-{\sin }^{2}C=\sin (B+C)\sin (B-C)$.

Also, since $A+B+C=180^{\circ }$, we know $\sin A=\sin (180^{\circ }-(B+C))=\sin (B+C)$. Substituting these in:

Dividing by $\sin (B+C)$, we arrive at the stunning conclusion: $\sin (B-C)=1$. This implies $B-C=90^{\circ }$.

With $B-C=90^{\circ }$ and $C=23^{\circ }$, we find:

We started with a complex algebraic expression and ended with a clear, geometric truth. This is the essence of JEE Advanced mathematics: taking the chaos of the problem and finding the elegant, simple order hidden underneath.