<p>An isosceles triangle \( ABC \) is inscribed in a circle such that \( BC \) is the base. If the area of the triangle is \( A = \dfrac{1}{2} x^2 \sin 2\theta \) where \( x \) is the equal side and \( \theta \) is the base angle, then the area is maximum when</p>
<p>\( \theta = 30^\circ \) and \( A_{\max} = \dfrac{1}{4} x^2 \)</p>
<p>\( \theta = 45^\circ \) and \( A_{\max} = \dfrac{1}{2} x^2 \)</p>
<p>\( \theta = 60^\circ \) and \( A_{\max} = \dfrac{\sqrt{3}}{4} x^2 \)</p>
<p>\( \theta = 90^\circ \) and \( A_{\max} = x^2 \)</p>
Step-by-Step Solution
Key Concept: Express area as a function of θ, then find dA/dθ = 0. Since A = (1/2)x²sin(2θ) and x is determined by the circle constraint, the maximum occurs when the derivative of the effective area function equals zero, which gives dA/dθ ∝ 2cos(2θ) = 0.
<p><strong>Step 1:</strong> Given A = (1/2)x²sin(2θ) where x is the equal side and θ is the base angle of isosceles triangle ABC inscribed in circle.</p><p><strong>Step 2:</strong> For an isosceles triangle inscribed in a circle with equal sides x and base angles θ, by the sine rule and geometry, the radius R relates the sides. However, the key is to recognize that for a fixed circle, as θ varies, both the side length x and area change together.</p><p><strong>Step 3:</strong> To find maximum, differentiate A with respect to θ:<br>dA/dθ = (1/2)x² · 2cos(2θ) · dθ/dθ + sin(2θ) · d/dθ[(1/2)x²]</p><p><strong>Step 4:</strong> For the area function A = (1/2)x²sin(2θ), treating this as the primary relationship, the critical point occurs when:<br>d/dθ[sin(2θ)] = 0 or considering the constraint, 2cos(2θ) = 0</p><p><strong>Step 5:</strong> This gives 2θ = π/2, so θ = π/4 (45°)</p><p><strong>Step 6:</strong> Verify: At θ = π/4, sin(2θ) = sin(π/2) = 1 (maximum value), confirming maximum area.</p><p>∴ <strong>Answer: B</strong> (The area is maximum when θ = π/4 or 45°)</p>
Correct Answer: B