Animated Solution for Mathematics - Trigonometry: Let $ABC$ and $AB{C}^{\prime }$ be two non-congruent triangles with sides $AB=4,AC=A{C}^{\prime }=2\sqrt{2}$ and angle $B=30^{\circ }$. The absolute value of the difference between the areas of these triangles is

A Journey into the SSA Case

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are dissecting a classic geometric trap.

The problem presents us with two triangles, $\Delta ABC$ and $\Delta AB{C}^{\prime }$, sharing side $AB=4$, side $AC=A{C}^{\prime }=2\sqrt{2}$, and angle $\angle B=30^{\circ }$. At first glance, this might seem like a standard trigonometry exercise, but it is actually a beautiful demonstration of the 'Ambiguous Case' in geometry.

The Visual Trap

Imagine you are standing at vertex $B$. You look out along a ray at an angle of $30^{\circ }$. You are told that vertex $A$ is at a distance of $4$ units from $B$.

Now, you are told that the distance from $A$ to the third vertex $C$ is $2\sqrt{2}$. If you were to draw a circle centered at $A$ with a radius of $2\sqrt{2}$, you would see it intersect the ray from $B$ at two distinct points.

These points are $C$ and $C^{\prime }$. This is why we have two non-congruent triangles! The side $AC$ is 'swinging' like a pendulum, creating two different possible configurations.

The Sine Rule as Our Compass

To navigate this, we need a tool that relates sides and angles. The Sine Rule is our best friend here. It states that for any triangle, the ratio of a side to the sine of its opposite angle is constant.

So, we write:

Substituting our known values, $AC=2\sqrt{2}$, $AB=4$, and $\angle B=30^{\circ }$, we get:

Since $\sin 30^{\circ }=\frac{1}{2}$, the left side becomes $\frac{2\sqrt{2}}{1/2}=4\sqrt{2}$. Thus, we have:

Solving for $\sin C$, we find $\sin C=\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}$.

The Two Faces of Angle C

Here is where the magic happens. We know that $\sin C=\frac{1}{\sqrt{2}}$. In the range $(0,180^{\circ })$, there are two angles that satisfy this: $45^{\circ }$ and $135^{\circ }$.

This confirms our geometric intuition! One triangle has an acute angle $C=45^{\circ }$, and the other has an obtuse angle $C^{\prime }=135^{\circ }$.

The Elegant Shortcut

The question asks for the absolute difference between the areas of $\Delta ABC$ and $\Delta AB{C}^{\prime }$. Instead of calculating the area of each triangle separately—which would involve finding the base $BC$ and $B{C}^{\prime }$—let's look at the geometry.

The difference between the two triangles is simply the area of the triangle $\Delta AC{C}^{\prime }$.

Since $AC=A{C}^{\prime }=2\sqrt{2}$, $\Delta AC{C}^{\prime }$ is an isosceles triangle. We know the exterior angle at $C^{\prime }$ is $135^{\circ }$, so the interior angle $\angle A{C}^{\prime }C=180^{\circ }-135^{\circ }=45^{\circ }$.

Because it is isosceles, $\angle AC{C}^{\prime }=45^{\circ }$ as well. The vertex angle $\angle CA{C}^{\prime }$ must be $180^{\circ }-(45^{\circ }+45^{\circ })=90^{\circ }$.

The Final Victory

We have discovered that $\Delta AC{C}^{\prime }$ is a right-angled isosceles triangle with legs of length $2\sqrt{2}$. The area of a right triangle is simply $\frac{1}{2}\times \text{base}\times \text{height}$.

Calculating this, we get $\frac{1}{2}\times 8=4$.

And there it is! The absolute difference in areas is 4. This problem teaches us that in the ambiguous case, the difference in areas is always the area of the isosceles triangle formed by the two possible positions of the third vertex.