Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If the sides of a right-angled triangle are in A.P., then the sines of the acute angles are</p>
<p>\(\dfrac{3}{5}, \dfrac{4}{5}\)</p>
<p>\(\dfrac{1}{\sqrt{3}}, \sqrt{\dfrac{2}{3}}\)</p>
<p>\(\dfrac{1}{2}, \dfrac{\sqrt{3}}{2}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: If sides are in A.P. with common difference d, use the Pythagorean theorem to find the ratio of sides, then calculate sines of acute angles from the resulting triangle ratios.
<p><strong>Step 1:</strong> Let the sides of the right-angled triangle in A.P. be (a-d), a, (a+d) where a > d > 0.</p><p><strong>Step 2:</strong> The longest side (a+d) is the hypotenuse. By Pythagorean theorem:<br/>(a-d)² + a² = (a+d)²</p><p><strong>Step 3:</strong> Expanding:<br/>a² - 2ad + d² + a² = a² + 2ad + d²<br/>2a² - 2ad + d² = a² + 2ad + d²<br/>a² = 4ad<br/>a = 4d</p><p><strong>Step 4:</strong> The sides are 3d, 4d, and 5d (ratio 3:4:5).</p><p><strong>Step 5:</strong> For the acute angles in a right triangle with sides 3, 4, 5:<br/>sin(α) = 3/5 and sin(β) = 4/5<br/>(where α and β are the two acute angles)</p><p><strong>Step 6:</strong> The sines of acute angles are <strong>√5/5 and 2√5/5</strong> (or equivalently <strong>1/√5 and 2/√5</strong>)</p><p>∴ Answer: A</p>
Correct Answer: A