Question

The angle of elevation of the top of a tower from a point A on the ground is ${{30}^{\circ }}$. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to ${{60}^{\circ }}$. Find the height of the tower and the distance of the tower from the point A.\[\]

Answer 1

Hint: Draw an equivalent figure of right angled triangle DAC where D is top of the tower, C is the bottom, B and A are two points on the ground. Then use the tangent ratio of the angles $\angle DAC,\angle DBC$ in the right-angled triangles DBC and DAC. You will get two equations of the height and distance of the tower. Solve it to find the asked value.

Complete step by step answer:

We draw a diagram to visualize the process. Here CD represents the tower where C is foot of the tower and D is top of the tower. As given in the question initially the measurement of angle elevation was done at point A and the angle was found to be $\angle DAC={{30}^{\circ }}$. Then after moving 20m towards the foot of the tower again measuring was done at the point B and the angle of elevation was found to be $\angle DBC={{60}^{\circ }}$ . \[\]
The distance between from the foot of the tower to the point A is $b$ then the distance to the point B is $\left( b-20 \right)m$. The height of the tower is denoted as $h=CD$\[\]
The triangles formed by $\Delta DAC,\Delta DBA$ are right-angled triangles. So the angles in them will follow trigonometric ratios.
In triangle $\Delta DAC$taking tangent of the angle $\angle DAC$ we get,
\[\begin{align}
  & \tan {{30}^{\circ }}=\dfrac{h}{b} \\
 & \Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{h}{b} \\
 & \Rightarrow b=h\sqrt{3}...\left( 1 \right) \\
\end{align}\]

Similarly, taking tangent of the angle $\angle DBC$ in triangle $\Delta DBC$ we get,
\[\begin{align}
  & \tan {{60}^{\circ }}=\dfrac{h}{b} \\
 & \Rightarrow \sqrt{3}=\dfrac{h}{b-20}...\left( 2 \right) \\
\end{align}\]
We put the value of $b$ from equation (1) in equation (2) to solve the pair of linear equations (1) and (2) and get,
\[\begin{align}
  & \sqrt{3}=\dfrac{h}{h\sqrt{3}-20} \\
 & \Rightarrow h=3h-20\sqrt{3} \\
 & \Rightarrow h=10\sqrt{3} \\
\end{align}\]

So the distance of the point A from the foot of tower is $10\sqrt{3}$.

Note: It is required in question to formulate relation between the perpendicular p=DC and the base b=BC or AC. That is why we used tangent trigonometric ratio$\left( \tan \theta =\dfrac{p}{b} \right)$. If we are asked to formulate relation between hypotenuse(h) and perpendicular we will use sine ratio$\left( \sin \theta =\dfrac{p}{h} \right)$ and cosine ratio for the relation between hypotenuse and perpendicular $\left( \cos \theta =\dfrac{b}{h} \right)$.