Animated Solution for Mathematics - Trigonometry: In a triangle $ABC$, medians $AD$ and $BE$ are drawn. If $AD=4$, $\angle DAB=\frac{\pi }{6}$ and $\angle ABE=\frac{\pi }{3}$, then the area of the $\Delta ABC$ is

Imagine you are standing in the middle of a triangle, $ABC$. You have drawn two medians, $AD$ and $BE$, which meet at a point $P$.

This point $P$ is the centroid, the center of mass, and the balancing point of the entire structure. In geometry, the centroid carries a secret: it divides every median in a $2:1$ ratio.

Specifically, for our median $AD$, the centroid $P$ splits it such that:

Since we are given $AD=4$, we immediately find that $AP=\frac{8}{3}$. This is our first foothold in this problem.

Now, let us turn our attention to the smaller triangle, $△ABP$. We are given $\angle PAB=30^{\circ }$ and $\angle PBA=60^{\circ }$.

If you sum these up, you get $90^{\circ }$. Since the sum of angles in any triangle is $180^{\circ }$, the third angle, $\angle APB$, must be:

This is a revelation! It means that $BP$ is perpendicular to $AD$. We have just discovered that the median $AD$ and the segment $BP$ are orthogonal.

In the right-angled triangle $△ABP$, we can now use simple trigonometry to find the length of $BP$. We know that:

Therefore, $BP=AP\times \tan (30^{\circ })$. Substituting our known values:

We are now ready to calculate the area. We want the area of $△ABC$, but let us start with $△ABD$.

The area of a triangle is $\frac{1}{2}\times \text{base}\times \text{height}$. For $△ABD$, we can treat $AD$ as the base and $BP$ as the height, because we just proved $BP\perp AD$.

So, the area of $△ABD$ is:

Finally, we recall the fundamental property of medians: a median divides a triangle into two triangles of equal area. Since $AD$ is a median of $△ABC$, the area of $△ABC$ is exactly twice the area of $△ABD$.

By rationalizing the denominator, we obtain the final result: