<p>The angles A, B and C of a triangle ABC are in arithmetic progression. If \(2b^2 = 3c^2\) then the angle A is:</p>
Step-by-Step Solution
Key Concept: Since angles A, B, C are in AP and sum to 180°, we can express them in terms of a common difference. Then use the Law of Sines with the given condition 2b² = 3c² to find angle A.
<p><strong>Step 1:</strong> Since angles A, B, C are in arithmetic progression, let them be (B - d), B, (B + d) for some common difference d.</p><p><strong>Step 2:</strong> Use the angle sum property: (B - d) + B + (B + d) = 180°<br>This gives: 3B = 180°, so B = 60°</p><p><strong>Step 3:</strong> Now apply the Law of Sines: b/sin B = c/sin C<br>Therefore: b/sin 60° = c/sin C<br>This gives: b = c · sin 60°/sin C = c · (√3/2)/sin C</p><p><strong>Step 4:</strong> Use the given condition 2b² = 3c²<br>So: 2[c · (√3/2)/sin C]² = 3c²<br>2 · c² · (3/4)/sin²C = 3c²<br>(3/2)/sin²C = 3<br>sin²C = 1/2<br>sin C = 1/√2, so C = 45°</p><p><strong>Step 5:</strong> Find angle A using the angle sum property:<br>A + B + C = 180°<br>A + 60° + 45° = 180°<br>A = 75°</p><p><strong>Step 6:</strong> Verify: If A = 75°, B = 60°, C = 45°, then B - A = -15° and C - B = -15°, confirming AP (in reverse). Also check 2b²/3c² using Law of Sines to confirm our answer.</p><p><strong>∴ Answer:</strong> b</p>
Correct Answer: b