<p>If in a triangle <i>ABC</i>, the altitudes from the vertices <i>A</i>, <i>B</i>, <i>C</i> on opposite sides are in HP, then sin <i>A</i>, sin <i>B</i>, sin <i>C</i> are in</p>
Step-by-Step Solution
Key Concept: Since altitudes are in HP, their reciprocals (which are proportional to the sides via area formula) are in AP. Use the relationship: altitude = 2Δ/side, where Δ is the area, to convert the HP condition into a relationship between sides, then apply sine rule.
<p><strong>Step 1:</strong> Let altitudes from A, B, C be h_A, h_B, h_C respectively. Given: h_A, h_B, h_C are in HP.</p><p><strong>Step 2:</strong> If h_A, h_B, h_C are in HP, then 1/h_A, 1/h_B, 1/h_C are in AP.</p><p><strong>Step 3:</strong> Using area formula: h_A = 2Δ/a, h_B = 2Δ/b, h_C = 2Δ/c, where a, b, c are sides opposite to angles A, B, C.</p><p><strong>Step 4:</strong> Therefore: 1/h_A = a/(2Δ), 1/h_B = b/(2Δ), 1/h_C = c/(2Δ)</p><p><strong>Step 5:</strong> Since 1/h_A, 1/h_B, 1/h_C are in AP, we have a, b, c are in AP.</p><p><strong>Step 6:</strong> By sine rule: a/sin A = b/sin B = c/sin C = 2R</p><p><strong>Step 7:</strong> Since a, b, c are in AP, and a = 2R sin A, b = 2R sin B, c = 2R sin C, we get sin A, sin B, sin C are in AP.</p><p>∴ Answer: <strong>B (AP)</strong></p>
Correct Answer: B