ABC is a triangle. Forces P, Q, R acting along IA, IB, and IC respectively are in equilibrium, where

Q: Why can't we just use vector components? It seems more robust than Lami's Theorem.

You absolutely can! Resolving forces along the $x$ and $y$ axes is a universal method that works for any system. However, in geometry-heavy problems like this, Lami's Theorem is the 'shortcut' that saves you precious minutes during the exam. It is about choosing the right tool for the job. Mastering Lami's Theorem allows you to bypass complex algebraic resolution and get straight to the geometric core of the problem.

Q: I always get confused with the angle $\angle BIC$. Is it always $90^\circ + A/2$?

Yes, it is a standard property of the incenter. Think of it as a 'geometric invariant' for any triangle. Whenever you see an incenter, that $90^\circ + A/2$ relationship should pop into your mind immediately. It is one of those 'JEE-essential' derivations that you should memorize or be able to derive in seconds.

Q: Why do we use $\sin(90^\circ + \theta) = \cos \theta$ here? Is there a physical reason?

It is purely a trigonometric simplification. The physics is contained within the Lami's equation; the math is just cleaning it up to match the options provided. Always look at your options—they are your roadmap! If the options were in terms of sine, we might have stopped earlier, but since they are in cosine, the allied angle identity is our bridge to the final answer.

Question

ABC is a triangle. Forces $\vec{P}, \vec{Q}, \vec{R}$ acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of $	riangle ABC$. Then $P : Q : R$ is

Options:

(A) $\sin A : \sin B : \sin C$

(B) $\sin \frac{A}{2} : \sin \frac{B}{2} : \sin \frac{C}{2}$

(C) $\cos \frac{A}{2} : \cos \frac{B}{2} : \cos \frac{C}{2}$

(D) $\cos A : \cos B : \cos C$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the elegant intersection of Euclidean geometry and classical mechanics.

Imagine you are standing at the incenter, $I$, of a triangle $	riangle ABC$, where the internal angle bisectors converge. You are holding three ropes, each exerting a force pulling outwards: $\vec{P}$ towards vertex $A$, $\vec{Q}$ towards vertex $B$, and $\vec{R}$ towards vertex $C$.

The system is in perfect equilibrium. We aim to determine how the magnitudes of these forces relate to the geometry of the triangle.

### The Geometry of the Incenter

The incenter $I$ is the center of the inscribed circle. By definition, the lines $IA$, $IB$, and $IC$ are the angle bisectors of $\angle A$, $\angle B$, and $\angle C$, respectively.

To understand the forces, we must first determine the angles between them: $\angle BIC$, $\angle CIA$, and $\angle AIB$. These angles define the spatial distribution of our force vectors.

### The Power of Lami's Theorem

When three concurrent forces are in equilibrium, **Lami's Theorem** is the most effective tool. It states that the magnitude of each force is proportional to the sine of the angle between the other two forces.

Mathematically, this is expressed as:

$$\frac{P}{\sin(\angle BIC)} = \frac{Q}{\sin(\angle CIA)} = \frac{R}{\sin(\angle AIB)}$$

This equation serves as the bridge between the physical world of forces and the geometric properties of the triangle.

### The Angle Derivation

Consider $	riangle BIC$. The sum of angles in any triangle is $180^{\circ}$. Since $IB$ and $IC$ are angle bisectors, we have $\angle IBC = \frac{B}{2}$ and $\angle ICB = \frac{C}{2}$.

Therefore, the angle at the incenter is:

$$\angle BIC = 180^{\circ} - \left(\frac{B}{2} + \frac{C}{2}ight)$$

In the main triangle $	riangle ABC$, we know $A + B + C = 180^{\circ}$, which implies $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$. Thus, $\frac{B}{2} + \frac{C}{2} = 90^{\circ} - \frac{A}{2}$.

Substituting this into our expression for $\angle BIC$, we obtain:

$$\angle BIC = 180^{\circ} - \left(90^{\circ} - \frac{A}{2}ight) = 90^{\circ} + \frac{A}{2}$$

By symmetry, the other angles are:

$$\angle CIA = 90^{\circ} + \frac{B}{2}$$

$$\angle AIB = 90^{\circ} + \frac{C}{2}$$

### Final Synthesis

Substituting these angles back into the Lami's Theorem equation, we get:

$$\frac{P}{\sin(90^{\circ} + \frac{A}{2})} = \frac{Q}{\sin(90^{\circ} + \frac{B}{2})} = \frac{R}{\sin(90^{\circ} + \frac{C}{2})}$$

Applying the trigonometric identity $\sin(90^{\circ} + 	heta) = \cos 	heta$, the denominators transform into cosines:

$$\frac{P}{\cos \frac{A}{2}} = \frac{Q}{\cos \frac{B}{2}} = \frac{R}{\cos \frac{C}{2}}$$

The final relationship between the magnitudes of the forces is:

**$$P : Q : R = \cos \frac{A}{2} : \cos \frac{B}{2} : \cos \frac{C}{2}$$**

This result demonstrates the inherent symmetry between the physical equilibrium of forces and the geometric properties of the triangle.

Animated Solution for Mathematics - Vector Algebra: ABC is a triangle. Forces $P, Q, R$ acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of $△ A B C$ . Then $P : Q : R$ is

Select Answer:

Visualized Solution

The Triangle and Incenter

Forces in Equilibrium

Lami's Theorem

Analyzing $△ B I C$

Sum of Angles in $△ B I C$

Using the Main Triangle Properties

Calculating $∠ B I C$

Finding the Other Angles

Substituting into Lami's Theorem

Trigonometric Simplification

Final Ratio

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Geometry of the Incenter

The Power of Lami's Theorem

The Angle Derivation

Final Synthesis

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