Multiple choice

The top of a pole, placed against a wall at an angle $\alpha$ with the horizon just touches the coping and when its foot is moved a meter away from the wall and its angle of inclination is $\beta$, it rests on the sill of a window, the vertical distance of the sill from the coping is

  1. $a \sin\left ( \dfrac{\alpha +\beta }{2} \right )$
  2. $-a \cot \left ( \dfrac{\alpha +\beta }{2} \right )$
  3. $a \cot \left ( \dfrac{\alpha +\beta }{2} \right )$
  4. $a \tan \left ( \dfrac{\alpha +\beta }{2} \right )$
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C Correct answer
Explanation

This is a classic trigonometry problem involving a ladder or pole leaning against a wall, where the vertical displacement is calculated using the difference in heights at two different angles.

AI explanation

Let the height of the wall coping from the ground be y and the horizontal distance from the wall to the original foot of the pole be x. This gives y equal to (x plus a) times tan alpha and also y equal to x times tan beta. Equating the expressions gives x times tan beta equal to x times tan alpha plus a times tan alpha, meaning x equals a times tan alpha divided by (tan beta minus tan alpha). The vertical distance of the sill from the coping is the difference between the total height and the window height, giving a distance of a multiplied by the sine of alpha times sine of beta divided by the sine of (beta minus alpha). Using trigonometric product to sum identities, this result simplifies exactly to a multiplied by the cotangent of the quantity (alpha plus beta) divided by 2.

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