Trigonometry & Inverse Trigonometry
Triangle Properties
Grade 11
Question:
<p>If A represents the area of acute angled triangle ABC, then \(\sqrt{a^2b^2 - 4A^2} + \sqrt{b^2c^2 - 4A^2} + \sqrt{c^2a^2 - 4A^2}\) is equal to</p>
<p>(a) \(a^2 + b^2 + c^2\)</p>
<p>(b) \(\frac{a^2 + b^2 + c^2}{2}\)</p>
<p>(c) ab\(\cos C + bc\cos A + ca\cos B\)</p>
<p>(d) ab\(\sin C + bc\sin A + ca\sin B\)</p>
Step-by-Step Solution
Key Concept: Use the area formula to eliminate the squared area terms and express them as products of sides with cosines.
<p>Using \(A = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B\):</p><p>\(a^2b^2 - 4A^2 = a^2b^2 - a^2b^2\sin^2 C = a^2b^2\cos^2 C\)</p><p>So \(\sqrt{a^2b^2 - 4A^2} = ab\cos C\)</p><p>Similarly for the other terms. Therefore the sum equals \(ab\cos C + bc\cos A + ca\cos B\).</p>
Correct Answer: C