Multiple choice

The angle of elevation of the top of a tower from a point A on the ground is $30^0$. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to $60^0$. The height of the tower is :

  1. $\sqrt{3}m$
  2. $5\sqrt{3}m$
  3. $10\sqrt{3}m$
  4. $20\sqrt{3}m$
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C Correct answer
Explanation

Let h be the height. From the 60 degree angle, the distance to the foot is h/sqrt(3). From the 30 degree angle, the distance is h/tan(30) = h*sqrt(3). The difference is 20m, so h*sqrt(3) - h/sqrt(3) = 20. Solving gives h(2/sqrt(3)) = 20, so h = 10*sqrt(3).

AI explanation

Let the height of the tower be H and the initial distance from point A to the foot of the tower be d. From point A, tan(30 degrees) = H / d, which gives the equation 1 divided by the square root of 3 equals H / d, so d equals H times the square root of 3. Point B is 20 meters closer to the tower, so the distance from point B to the foot is d minus 20. From point B, tan(60 degrees) = H / (d - 20), and since tan(60 degrees) equals the square root of 3, we write the square root of 3 equals H / (d - 20). Substituting d from the first equation gives the square root of 3 equals H / (H times the square root of 3 minus 20). Cross-multiplying yields 3 times H minus 20 times the square root of 3 equals H. Subtracting H from both sides gives 2H equals 20 times the square root of 3. Dividing by 2 gives the height of the tower H as 10 times the square root of 3 meters.

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