Question

On a horizontal plane, there is a vertical tower with a flagpole on the top of the tower. At a point 9 meters away from the foot of the tower the angle of elevation of the top and bottom of the flagpole are \[{{30}^{\circ }}\] and \[{{60}^{\circ }}\] respectively. Find the height of the tower and the flagpole mounted on it.

Answer 1

Hint: Consider the maximum height as AC 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 heights.

Complete step-by-step answer:

In the figure flag pole is AB, BC is tower and CD is base.

In \[\Delta ACD\]

\[\tan {{60}^{\circ }}=\dfrac{AC}{DC}\]

\[\sqrt{3}=\dfrac{AC}{9}\]

\[AC=9\sqrt{3}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)

Now in \[\Delta DBC\]

\[\tan {{30}^{\circ }}=\dfrac{BC}{DC}\]

\[\dfrac{1}{\sqrt{3}}=\dfrac{BC}{9}\]

\[BC=\dfrac{9}{\sqrt{3}}=3\sqrt{3}m\]

To find the height of flagpole,
Substituting (1),

\[AC=AB+BC\]

\[AB=AC-BC\]

\[AB=9\sqrt{3}-3\sqrt{3}=6\sqrt{3}m\]

Height of the flagpole is \[6\sqrt{3}m\] and that of the tower is \[3\sqrt{3}\]m.

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.