Three forces P, Q and R acting along IA, IB and IC, where I is the incentre of a ABC, are in equilibrium. Then P:Q:R is - Mathematics MCQ AIEEE 2004 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

The angles at the incentre I between the segments I A, I B, and I C can be determined. The angle ∠ B I C = 18 0^{\circ} - (∠ B /2 + ∠ C /2) = 18 0^{\circ} - (18 0^{\circ} - ∠ A) /2 = 9 0^{\circ} + ∠ A /2 . Similarly, ∠ A I C = 9 0^{\circ} + ∠ B /2 and ∠ A I B = 9 0^{\circ} + ∠ C /2 . By Lami's theorem for forces in equilibrium: \frac{P}{s i n ∠ B I C} = \frac{Q}{s i n ∠ A I C} = \frac{R}{s i n ∠ A I B} P : Q : R = sin (9 0^{\circ} + A /2) : sin (9 0^{\circ} + B /2) : sin (9 0^{\circ} + C /2) . Since sin (9 0^{\circ} + θ) = cos θ : P : Q : R = cos (A /2) : cos (B /2) : cos (C /2) .

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

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

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

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

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

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