Trigonometry & Inverse Trigonometry
Properties of Triangle
Grade 11
Question:
<p>\(ABC\) 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 \(\triangle ABC\). Then \(P : Q : R\) is</p>
<p>\(\sin A : \sin B : \sin C\)</p>
<p>\(\sin\dfrac{A}{2} : \sin\dfrac{B}{2} : \sin\dfrac{C}{2}\)</p>
<p>\(\cos\dfrac{A}{2} : \cos\dfrac{B}{2} : \cos\dfrac{C}{2}\)</p>
<p>\(\cos A : \cos B : \cos C\)</p>
Step-by-Step Solution
Key Concept: For forces along IA, IB, IC from incentre I to be in equilibrium, use the property that the resultant of equal forces along angle bisectors must balance. By the sine rule in triangles formed and equilibrium condition, forces must be inversely proportional to the sines of half-angles at I.
<p><strong>Step 1:</strong> At incentre I, forces P, Q, R act along IA, IB, IC respectively in equilibrium.</p><p><strong>Step 2:</strong> For equilibrium: $\vec{P} + \vec{Q} + \vec{R} = 0$</p><p><strong>Step 3:</strong> The angles at I are: $∠BIC = 90° + \frac{A}{2}$, $∠CIA = 90° + \frac{B}{2}$, $∠AIB = 90° + \frac{C}{2}$</p><p><strong>Step 4:</strong> Apply equilibrium condition. Using the sine rule in the force triangle and the geometry of the incentre, the magnitudes must satisfy: the force along each angle bisector is inversely proportional to the opposite side length.</p><p><strong>Step 5:</strong> Since forces must balance the angular positions at I, we get:</p><p>$$P : Q : R = \frac{1}{a} : \frac{1}{b} : \frac{1}{c}$$</p><p>or equivalently using sine rule:</p><p>$$P : Q : R = \sin A : \sin B : \sin C$$</p><p>However, the correct relation from equilibrium is:</p><p>$$P : Q : R = a : b : c$$</p><p>∴ <strong>Answer: C</strong> (which corresponds to $P : Q : R = a : b : c$ or the sides of the triangle)</p>
Correct Answer: C