Trigonometry & Inverse Trigonometry
Properties of Triangle Angles
Grade 11
Question:
<p>The product of the sines of the angles of a triangle is \(p\) and the product of their cosines is \(q\). Then, the tangents of the angles are the roots of the equation</p>
<p>(a) \(qx^3 - px^2 + (1 + q)x - p = 0\)</p>
<p>(b) \(qx^3 - px^2 - (1 - q)x - p = 0\)</p>
<p>(c) \(qx^3 - px^2 + (1 + q)x + p = 0\)</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Use the property that tangents of angles in a triangle satisfy \(\tan A + \tan B + \tan C = \tan A \tan B \tan C\) and relate to the given products of sines and cosines.
<p>Let \(\tan A, \tan B, \tan C\) be the roots.</p><p>For a triangle: \(\tan A + \tan B + \tan C = \tan A \tan B \tan C\)</p><p>Also: \(\sin A \sin B \sin C = p\) and \(\cos A \cos B \cos C = q\)</p><p>From Vieta's formulas for the cubic equation, we derive: \(qx^3 - px^2 + (1 + q)x - p = 0\)</p>
Correct Answer: A