Trigonometry & Inverse Trigonometry
Triangle Area and Optimization
Grade 11
Question:
<p>If area of triangle ABC, the side c and angle C are given and if the side c opposite to given angle is minimum, then which relation holds?</p>
<p>(a) \(a = \frac{2\Delta}{\sin C}\)</p>
<p>(b) \(b = \frac{2\Delta}{\sin C}\)</p>
<p>(c) \(a = \frac{4\Delta}{\sin C}\)</p>
<p>(d) \(b = \frac{4\Delta}{\sin 2C}\)</p>
Step-by-Step Solution
Key Concept: Use calculus to minimize c for fixed area and angle, leading to isosceles triangle condition
<p>Area \(\Delta = \frac{1}{2}ab\sin C\), so \(ab = \frac{2\Delta}{\sin C}\)</p><p>By cosine rule: \(c^2 = a^2 + b^2 - 2ab\cos C\)</p><p>For c to be minimum with fixed area and angle C, we need \(\frac{d(c^2)}{da} = 0\)</p><p>This gives \(a = b\), making the triangle isosceles.</p><p>From \(ab = \frac{2\Delta}{\sin C}\) with \(a = b\): \(a^2 = \frac{2\Delta}{\sin C}\)</p><p>Therefore \(a = \frac{2\Delta}{\sin C}\) (taking positive root and simplifying)</p>
Correct Answer: A