Question:

<p><strong>Step 1:</strong> Since angles <strong>A</strong>, <strong>B</strong>, <strong>C</strong> are in AP, let them be <strong>B</strong> - <strong>d</strong>, <strong>B</strong>, <strong>B</strong> + <strong>d</strong>.</p><p><strong>Step 2:</strong> From <span>$A + B + C = 180°$</span>: <span>$(B-d) + B + (B+d) = 180°$</span></p><p>Therefore: <span>$3B = 180° \Rightarrow B = 60°$</span></p><p><strong>Step 3:</strong> Use the sine rule: <span>$\frac{a}{\sin A} = \frac{b}{\sin B}$</span></p><p>Given <span>$a : b = 1 : \sqrt{3}$</span>, so <span>$\frac{a}{b} = \frac{1}{\sqrt{3}} = \frac{\sin A}{\sin B}$</span></p><p><strong>Step 4:</strong> With <span>$B = 60°$</span> and <span>$\sin B = \frac{\sqrt{3}}{2}$</span>: <span>$\sin A = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2}$</span></p><p>Therefore: <span>$A = 30°$</span>, which gives <span>$C = 90°$</span></p><p><strong>Step 5:</strong> Use the area formula: <span>$\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}ab\sin 90° = \frac{1}{2}ab$</span></p><p><strong>Step 6:</strong> From sine rule with <span>$c = 4$</span>: <span>$\frac{c}{\sin C} = \frac{4}{1} = 4 = 2R$</span></p><p>So <span>$a = 2\sin A = 2 \cdot \frac{1}{2} = 1$</span> and <span>$b = 2\sin B = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$</span></p><p><strong>Step 7:</strong> Area <span>$= \frac{1}{2} \cdot 1 \cdot \sqrt{3} = \frac{\sqrt{3}}{2} \approx 0.866$</span> OR using <span>$\Delta = \frac{1}{2}ac\sin B = \frac{1}{2} \cdot 1 \cdot 4 \cdot \sin 60° = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$</span></p><p>Actually, recalculating: <span>$a = 2, b = 2\sqrt{3}$</span> gives Area <span>$= \frac{1}{2} \cdot 2 \cdot 2\sqrt{3} \cdot \sin 90° = 2\sqrt{3}$</span></p><p>∴ Answer is (c) <span>2$\sqrt{3}$</span></p>