Question

The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is 32. When the observer moves towards the tower a distance of 100 in, he finds the angle of elevation of the top lobe 63. Find the height of the tower and the distance of the first position from the tower. (Take tan 32 = 0,624S and tan 63 = 1.9626)

Answer 1

Hint: consider the maximum height as AB and draw the angle of elevations at two different points and apply \[\tan \theta \]to the two right angled triangles and we will get two equations and then we have to compute the maximum height from which he falls.

Complete step-by-step answer:

Let \[CB=x\].

\[\tan {{32}^{\circ }}=\dfrac{AB}{BD}\]

\[0.6248=\dfrac{AB}{100+x}\]

\[62.48+0.6248x=AB\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

Now,

\[\tan {{63}^{\circ }}=\dfrac{AB}{CB}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)

(1) is substituted in (2)

\[1.9626=\dfrac{62.48+0.6248x}{x}\]

\[1.9626x-0.6248x=62.48\]

\[1.334x=62.48\]

\[x=\dfrac{62.48}{1.334}\]

\[x=46.829\]

Therefore AB is \[AB=62.48+(0.6248\times 46.8)\]

\[AB=91.72m\]

Note: The angle of elevation is the angle between the horizontal line from the observer and the line of sight to an object that is above the horizontal line. As the person moves from one point to another angle of elevation varies. If we move closer to the object the angle of elevation increases and vice versa.