Multiple choice

The distance between two vertical poles is $d\ metres$. The height of one of the poles is double the height of the other. The angles of elevation of the tops of the poles from the middle point of the line joining their feet are complementary to each other. The height of the shorter pole is

  1. $\dfrac {d}{\sqrt 2}$
  2. $\dfrac {d}{2}$
  3. $\dfrac {d}{4}$
  4. $\dfrac {d}{2\sqrt 2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let height of shorter pole be h, taller be 2h. Distance from middle point is d/2. tan(theta) = h / (d/2) = 2h/d. tan(90-theta) = cot(theta) = 2h / (d/2) = 4h/d. Since tan(theta) * cot(theta) = 1, (2h/d) * (4h/d) = 1. 8h^2 / d^2 = 1. h^2 = d^2 / 8. h = d / sqrt(8) = d / (2 * sqrt(2)).

AI explanation

Let the height of the shorter pole be h, so the taller pole is 2h, and the distance from the middle point to either pole is d/2. If the angles of elevation are complementary, say theta and (90 degrees minus theta), their tangents are reciprocals. Using the tangent ratio, h/(d/2) = tan(theta) and 2h/(d/2) = tan(90 degrees minus theta), which simplifies to 4h/d = 1/tan(theta). Substituting the first equation into the second gives 4h/d = (d/2)/h, so 4h^2 = d^2/4. Solving for h results in h = d/(2*sqrt(2)).

Keep practicing — related questions