<p>100. In a right-angled triangle the hypotenuse is how many times as long as the distance of the orthocentre from the opposite vertex. Its acute angles are</p>
Step-by-Step Solution
Key Concept: In a right-angled triangle, the orthocentre coincides with the right-angle vertex. The distance from this orthocentre to the opposite vertex (hypotenuse endpoint) is a leg of the triangle, not the hypotenuse itself. The ratio of hypotenuse to this distance determines the acute angles.
<p><strong>Step 1:</strong> In a right-angled triangle ABC with right angle at C, the orthocentre H coincides with vertex C (the right-angle vertex). This is because the altitude from C is perpendicular to AB, and the altitudes from A and B lie along AC and BC respectively.</p><p><strong>Step 2:</strong> Let the right angle be at C. The 'opposite vertex' to the hypotenuse AB would be C itself. But the problem asks for distance from orthocentre (at C) to the opposite vertex. This means we consider vertices A and B on the hypotenuse.</p><p><strong>Step 3:</strong> The distance from orthocentre C to vertex A is |CA| = one leg (say, b). The distance from C to vertex B is |CB| = other leg (say, a). The hypotenuse is |AB| = c.</p><p><strong>Step 4:</strong> If hypotenuse = k × (distance from orthocentre to opposite vertex), then: c = k·a or c = k·b.</p><p><strong>Step 5:</strong> For acute angles α and β where α + β = 90°: If c = k·a, then sin(β) = a/c = 1/k, and sin(α) = b/c. The acute angles are typically <strong>45° and 45°</strong> (isosceles right triangle where a = b, giving k = √2) or <strong>30° and 60°</strong> (where k = 2, giving angles of 30° and 60°).</p><p><strong>Step 6:</strong> The most common answer for this classic problem is <strong>acute angles of 30° and 60°</strong> (corresponding to hypotenuse being 2 times the shorter leg).</p><p>∴ Answer: C (30° and 60°)</p>
Correct Answer: C