Multiple choice

Tower is observed from two stations $A$and $B$ where $B$ is East of $A$ at a distance $100m$. The tower is to north of $A$ and to North-West of $B$. The angles of elevation of the tower from $A$and $B$ are complementary. The height of the tower (in meters) is:

  1. $100$ $\sqrt{2}$
  2. $100(2)^{1/4}$
  3. $100(2)^{-1/4}$
  4. $100(2)^{1/3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let h be the height of the tower at O. Let A be at (0,0,0) and B at (100,0,0). The tower is at (0, y, 0). From A, tan(alpha) = h/y. From B, the tower is at North-West, so the distance from B to the base is sqrt(100^2 + y^2). tan(90-alpha) = h/sqrt(100^2 + y^2). Thus, cot(alpha) = h/sqrt(100^2 + y^2). Multiplying tan(alpha) and cot(alpha) gives 1 = h^2 / (y * sqrt(100^2 + y^2)). Solving for h leads to h = 100 * 2^(1/4).

AI explanation

Let the tower height be h and its location be T. In the horizontal plane, triangle ABT is a right triangle at A with angle ABT equal to 45 degrees and AB equal to 100 m, making AT equal to 100 m. If the angle of elevation from A is theta, then h equals 100 tan theta. Since the elevations are complementary, the elevation from B is 90 minus theta, giving h equals 100 tan(90 minus theta) which equals 100 cot theta. Squaring and multiplying the two equations gives h squared equals 10000, so h equals 100(2) to the power of 1/4.

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