Trigonometry & Inverse Trigonometry
Half-Angle Formulas in Triangles
Grade 11
Question:
<p><strong>Ex. 24:</strong> <strong>Statement I</strong> In a triangle ABC, \(\cos^2\dfrac{A}{2}\) has the value equal to \(\dfrac{s(s-a)}{abc}\)</p><p><strong>Statement II</strong> In a triangle ABC, \(\cos\dfrac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}}\), \(\cos\dfrac{B}{2} = \sqrt{\dfrac{(s-a)(s-c)}{ac}}\), \(\cos\dfrac{C}{2} = \sqrt{\dfrac{(s-a)(s-b)}{ab}}\)</p>
<p>(a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.</p>
<p>(b) Statement I is True, Statement II is True; Statement II is NOT a correct explanation for Statement I.</p>
<p>(c) Statement I is True, Statement II is False.</p>
<p>(d) Statement I is False, Statement II is True.</p>
Step-by-Step Solution
Key Concept: The half-angle cosine formulas for a triangle involve the product of differences of semi-perimeter from two sides divided by the product of those sides, not involving the full semi-perimeter times its difference.
<p><strong>Step 1:</strong> Using Statement II formula: \(\cos\dfrac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}}\)</p><p><strong>Step 2:</strong> Therefore, \(\cos^2\dfrac{A}{2} = \dfrac{(s-b)(s-c)}{bc}\)</p><p><strong>Step 3:</strong> This does not equal \(\dfrac{s(s-a)}{abc}\) as stated in Statement I.</p><p><strong>Step 4:</strong> Statement II formulas are correct.</p><p>∴ Answer is (c).</p>
Correct Answer: c