Animated Solution for Mathematics - Straight Lines: Consider a triangle $ABC$ having the vertices $A(1,2)$, $B(\alpha ,\beta )$ and $C(\gamma ,\delta )$ and angles $\angle ABC=\frac{\pi }{6}$ and $\angle BAC=\frac{2\pi }{3}$. If the points $B$ and $C$ lie on the line $y=x+4$, then ${\alpha }^2+{\gamma }^2$ is equal to ________

A Journey into Triangle $ABC$

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are uncovering the hidden symmetry within a triangle.

When you first look at a problem like this, it is easy to feel overwhelmed by the variables $\alpha ,\beta ,\gamma ,$ and $\delta$. But take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

The Hidden Isosceles Nature

We begin with a point $A(1,2)$ and a line $BC$ defined by $y=x+4$. The points $B$ and $C$ are anchored to this line.

We are given $\angle BAC=120^{\circ }$ and $\angle ABC=30^{\circ }$. Before we touch a single algebraic equation, let us use the most powerful tool in geometry: the angle sum property. The sum of angles in any triangle is $180^{\circ }$.

Thus, $\angle ACB=180^{\circ }-(120^{\circ }+30^{\circ })=30^{\circ }$.

Stop for a moment and appreciate this. We have $\angle ABC=30^{\circ }$ and $\angle ACB=30^{\circ }$. This means our triangle is isosceles!

The sides $AB$ and $AC$ are equal. This symmetry is not just a geometric curiosity; it is the engine that will drive our calculation to a clean finish.

The Slope Strategy

Now, we need to find the lines $AB$ and $AC$. We know the slope of the base line $BC$ is $m=1$ (from $y=1x+4$).

We also know that lines $AB$ and $AC$ make an angle of $30^{\circ }$ with this base line. To find the slopes of $AB$ and $AC$, we use the angle between two lines formula:

Substituting $\theta =30^{\circ }$ and $m=1$, we get:

Solving this for $m^{\prime }$ gives us two distinct slopes: $m_1=2+\sqrt{3}$ and $m_2=2-\sqrt{3}$. These are the slopes of our two sides. It is fascinating how the irrationality of $\sqrt{3}$ appears here, but do not fear it. It is merely a placeholder for the geometry we are about to solve.

The Intersection and the Algebra

With the slopes in hand, we construct the lines. For line $AB$, passing through $A(1,2)$ with slope $m_1=2+\sqrt{3}$, the equation is $y-2=(2+\sqrt{3})(x-1)$.

We intersect this with $y=x+4$ to find point $B(\alpha ,\beta )$. Substituting $y=x+4$ into our line equation, we get:

This simplifies to $x+2=(2+\sqrt{3})x-(2+\sqrt{3})$. Solving for $x$, we find:

Rationalizing this, we multiply the numerator and denominator by $(\sqrt{3}-1)$, yielding:

We repeat this exact process for point $C(\gamma ,\delta )$ using the second slope $m_2=2-\sqrt{3}$. The algebra follows the same path, leading us to:

Rationalizing this, we obtain:

Cancellation

We are asked for ${\alpha }^2+{\gamma }^2$. Let us square these values:

Look at that! When we add them, the terms involving $\sqrt{3}$ are $-\frac{3\sqrt{3}}{2}$ and $+\frac{3\sqrt{3}}{2}$. They cancel out perfectly!

We are left with $7+7=14$.

This is the beauty of mathematics. We started with a complex geometric setup, navigated through slopes and irrational numbers, and arrived at a simple, elegant integer. The final answer is 14.