Multiple choice

True or False DE is a tower standing on a horizontal plane and ABCD is a straight line in the plane. The height of the tower substends an angle $\theta$ at A, 2$\theta$ at B and 3$\theta$ at C. If AB and BC be respectively 50 metres and 20 metres then the height of the tower and the distance CD are $\dfrac{25}{2} \sqrt{7} m$ and $17.5$ m.

  1. True

  2. False
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A Correct answer
Explanation

This is a classic trigonometry problem. Using the cotangent rule for the angles theta, 2theta, and 3theta at distances 50, 20, and x, the derived height and distance values match the provided statement.

AI explanation

In the triangle formed by the tower and the two segments, the exterior angle theorem gives 2 theta at B equal to theta at A plus the top angle, making the top angle equal to theta. At C, 3 theta at C equals theta at A plus the top angle at the further point, making the top angle 2 theta. Using the sine rule in the triangle with the 20 m and 50 m sides, the common segment factor evaluates to 25 divided by the square root of 7. The height of the tower is 20 sin theta plus 50 sin 2 theta, which simplifies exactly to 25 divided by 2 times the square root of 7 m, and the distance CD evaluates to 17.5 m, making the statement true.

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