Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>Two vertical poles of heights, 20 m and 80 m stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is __________ .</p>
Step-by-Step Solution
Key Concept: Use similar triangles formed by the intersecting lines. If two lines from top of pole 1 to foot of pole 2 and vice versa intersect, the height of intersection can be found using the property that the intersection point divides the connecting lines proportionally based on pole heights.
<p><strong>Step 1:</strong> Let the two poles have heights h₁ = 20 m and h₂ = 80 m, separated by distance d (which cancels out).</p><p><strong>Step 2:</strong> Line 1 goes from top of pole 1 (height 20) to foot of pole 2. Line 2 goes from top of pole 2 (height 80) to foot of pole 1.</p><p><strong>Step 3:</strong> At intersection point P with height h, by similar triangles: the ratio of distances on the horizontal plane from line 1 and line 2 satisfy: h/(20-h) = d₁/d and (80-h)/h = d₂/d where d₁ + d₂ = d.</p><p><strong>Step 4:</strong> From similar triangles: h/20 + h/80 = 1. This gives h(1/20 + 1/80) = 1.</p><p><strong>Step 5:</strong> h(4/80 + 1/80) = 1 → h(5/80) = 1 → h = 80/5 = 16.</p><p>Alternatively, using the formula: h = (h₁ × h₂)/(h₁ + h₂) = (20 × 80)/(20 + 80) = 1600/100 = 16.</p><p>∴ <strong>Answer: 16 meters</strong></p>
Correct Answer: 16