Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>From a point O on the ground, poles of equal heights are placed at equal distances \(k\) apart along a straight line. The angle of elevation from O to the top of the 10th pole is \(\alpha\). If the distance from O to the base of the first pole is \(a\), the height \(h\) of each pole is</p>
<p>(1) \(\dfrac{h\cos\alpha - a\sin\alpha}{9\sin\alpha}\)</p>
<p>(2) \(\dfrac{h\sin\alpha - a\cos\alpha}{9\cos\alpha}\)</p>
<p>(3) \(\dfrac{h\cos\alpha + a\sin\alpha}{9\sin\alpha}\)</p>
<p>(4) \(\dfrac{h\cos\alpha - a\sin\alpha}{9\cos\alpha}\)</p>
Step-by-Step Solution
Key Concept: Set up a coordinate system with O at origin; the nth pole is at horizontal distance a + (n-1)k from O. Use the angle of elevation condition for the 10th pole to relate h, a, k, and α through tangent.
<p><strong>Step 1:</strong> Set up the geometry. Point O is on the ground. The first pole's base is at horizontal distance <em>a</em> from O. Poles are spaced <em>k</em> apart, so the <em>n</em>th pole is at distance <em>a</em> + (<em>n</em>−1)<em>k</em> from O.</p><p><strong>Step 2:</strong> For the 10th pole, the horizontal distance from O to its base is: <em>a</em> + (10−1)<em>k</em> = <em>a</em> + 9<em>k</em></p><p><strong>Step 3:</strong> The angle of elevation α from O to the top of the 10th pole gives: tan(α) = <em>h</em>/(<em>a</em> + 9<em>k</em>)</p><p><strong>Step 4:</strong> Solve for <em>h</em>:</p><p><em>h</em> = (<em>a</em> + 9<em>k</em>) tan(α)</p><p>∴ Answer: <strong>h = (a + 9k) tan α</strong></p>
Correct Answer: A