Animated Solution for Mathematics - Trigonometry: In a triangle $ABC$, $\angle B=\pi /3$ and $\angle C=\pi /4$. Let $D$ divide $BC$ internally in the ratio $1:3$ then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

Unlocking the Triangle

Welcome, student. Today, we are not just solving a geometry problem; we are embarking on a journey of logical deduction. When you look at a triangle like $△ABC$, it is easy to see it as a static shape.

But in the world of JEE Advanced, a triangle is a dynamic system of relationships. We are given $\angle B=\frac{\pi }{3}$ and $\angle C=\frac{\pi }{4}$. We have a point $D$ on $BC$ that divides it in a ratio of $1:3$.

Our mission is to find the ratio of the sines of the two angles created by the cevian $AD$. Let us define our target angles as $\angle BAD=\alpha$ and $\angle CAD=\beta$. We are hunting for the value of $\frac{\sin \alpha }{\sin \beta }$.

The Art of Visualization

Imagine standing at vertex $A$ and looking down at the base $BC$. There is a line, a cevian $AD$, that cuts through the triangle. This line splits our large triangle into two smaller, distinct entities: $△ABD$ and $△ACD$.

We are told $D$ divides $BC$ in a ratio of $1:3$. If we define the length of $BD$ as some variable $x$, then $DC$ must be $3x$.

We do not know the absolute length of $BC$, and as we will soon see, we do not need to. The ratio is what matters.

The Bridge Strategy

How do we connect the left side of the triangle to the right side? The side $AD$ is the common wall between our two rooms, $△ABD$ and $△ACD$.

If we express the length of $AD$ in terms of the knowns in $△ABD$, and then again in terms of the knowns in $△ACD$, we can equate them. We invoke the Sine Rule, which states that the ratio of a side to the sine of its opposite angle is constant.

For $△ABD$:

For $△ACD$:

The Algebraic Dance

Since both expressions represent the same physical length $AD$, we equate them:

Substituting our knowns ($BD=x$, $DC=3x$, $\angle B=\frac{\pi }{3}$, and $\angle C=\frac{\pi }{4}$):

The variable $x$ appears on both sides and is non-zero, so we can safely cancel it. This confirms that the ratio is independent of the triangle's size. Rearranging to isolate our target ratio:

The Final Calculation

We know the standard trigonometric values:

Substituting these into our ratio:

Simplifying the expression:

Thus, the final ratio is $\frac{1}{\sqrt{6}}$.

We started with a triangle and a simple division of a base. By identifying the common side $AD$ as a bridge and applying the Sine Rule, we navigated through the geometry to find a precise ratio.

Remember this technique: whenever you see a triangle split by a cevian, look for the common side. It is the key that unlocks the door to the solution.