Trigonometry & Inverse Trigonometry
Trigonometric Applications
Grade 11
Question:
<p>Two parallel chords are drawn on the same side of the centre of a circle of radius <i>R</i>. It is found that they subtend an angle of <i>θ</i> and <i>2θ</i> at the centre of the circle. The perpendicular distance between the chords is</p>
<p>(a) <i>2R</i> sin(θ/2) sin(θ)</p>
<p>(b) <i>R</i>[1 - cos(θ/2)][1 + 2cos(θ/2)]</p>
<p>(c) <i>R</i>[1 + cos(θ/2)][1 - 2cos(θ/2)]</p>
<p>(d) <i>2R</i> sin(θ/4) sin(3θ/4)</p>
Step-by-Step Solution
Key Concept: Use the perpendicular distance from centre to chord formula (p = R cos α) where α is half the central angle, then find the difference between the two distances.
<p><strong>Step 1:</strong> For chord subtending angle θ at centre: OM = p₁ = R cos(θ/2)</p><p><strong>Step 2:</strong> For chord subtending angle 2θ at centre: ON = p₂ = R cos(θ)</p><p><strong>Step 3:</strong> Perpendicular distance MN = p₁ - p₂ = R[cos(θ/2) - cos(θ)]</p><p><strong>Step 4:</strong> Using cos(θ) = 2cos²(θ/2) - 1, we get:</p><p>MN = R[cos(θ/2) - (2cos²(θ/2) - 1)]</p><p>= R[1 - cos(θ/2)][1 + 2cos(θ/2)]</p><p>Converting using factorisation and trigonometric identities yields the answer.</p><p>∴ Answer is D.</p>
Correct Answer: D