ABC is a triangle. Forces P, Q, R acting along IA, IB and IC respectively are in equilibrium, where I is the incentre of ABC . Then P : Q : R is: - Mathematics MCQ AIEEE 2026 | ExamTest.live

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A BC is a triangle. Forces P, Q, R acting along I A, I B and I C respectively are in equilibrium, where I is the incentre of △ A BC . Then P : Q : R is:

Solution & Step-by-step Explanation

By Lami's Theorem, for equilibrium: \frac{P}{sin ( ∠ B I C )} = \frac{Q}{sin ( ∠ C I A )} = \frac{R}{sin ( ∠ A I B )} In △ A BC, the angles formed at the incentre I are: ∠ B I C = 18 0^{\circ} - (\frac{B}{2} + \frac{C}{2}) = 18 0^{\circ} - (9 0^{\circ} - \frac{A}{2}) = 9 0^{\circ} + \frac{A}{2} Similarly, ∠ C I A = 9 0^{\circ} + \frac{B}{2} and ∠ A I B = 9 0^{\circ} + \frac{C}{2} .Thus: \frac{P}{sin ( 9 0 ^{\circ} + A /2 )} = \frac{Q}{sin ( 9 0 ^{\circ} + B /2 )} = \frac{R}{sin ( 9 0 ^{\circ} + C /2 )} \frac{P}{cos ( A /2 )} = \frac{Q}{cos ( B /2 )} = \frac{R}{cos ( C /2 )} So, P : Q : R = cos \frac{A}{2} : cos \frac{B}{2} : cos \frac{C}{2} .

Try it yourself before checking the explanation above.

A BC is a triangle. Forces P, Q, R acting along I A, I B and I C respectively are in equilibrium, where I is the incentre of △ A BC . Then P : Q : R is:

sin A : sin B : sin C

sin \frac{A}{2} : sin \frac{B}{2} : sin \frac{C}{2}

cos \frac{A}{2} : cos \frac{B}{2} : cos \frac{C}{2}

cos A : cos B : cos C

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