Question

The angle of elevation of the top of a building from a point on the same level as the foot of the building is 30°. On advancing 150m towards the foot of the tower, the angle of elevation of the top of the building becomes 60°. Show that the height of the tower is 129.9m
$\sqrt{3}=1.73$

Answer 1

Hint: We are given angles of elevation at two points. We know that $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$ hence at each position we will take tan of the angle and form two equations and then we will solve the equation such that we get the height of tower.

Complete step by step answer:

Now the angle of elevation of the top of a building from a point on the same level as the foot of the building is 30°.
Consider the triangle made by the building and this point let us say ABC.
Now here ABC is a right angle triangle with AB as height and BC as base.
Now we know $\tan \theta =\dfrac{\text{opposite side}}{\text{adjecent side}}$
Hence we get $\tan {{30}^{\circ }}=\dfrac{AB}{BC}$
Now let BD be Xm then we have BC = 150m + Xm
Hence substituting the values of BC we get
$\tan 30=\dfrac{AB}{x+150}$
Now we know that the value of tan30°= $\dfrac{1}{\sqrt{3}}$
Hence we have
$\begin{align}
  & \dfrac{1}{\sqrt{3}}=\dfrac{AB}{x+150} \\
 & \Rightarrow x+150=\sqrt{3}AB \\
 & \Rightarrow x=\sqrt{3}AB-150.................(1) \\
\end{align}$
Now it is given that on advancing 150m towards the foot of the tower, the angle of elevation of the top of the building becomes 60°
Now here ABD is a right angle triangle with AB as height and BD as base.
Now we know $\tan \theta =\dfrac{\text{opposite side}}{\text{adjecent side}}$
Hence we get $\tan {{60}^{\circ }}=\dfrac{AB}{BD}$
Now we have assumed BD to be x and we know that the value of $\tan {{60}^{\circ }}=\sqrt{3}$ .
Hence we get
$\begin{align}
  & \sqrt{3}=\dfrac{AB}{x} \\
 & \Rightarrow x=\dfrac{AB}{\sqrt{3}}...........................(2) \\
\end{align}$
Hence from equation (1) and equation (2) we get.
$\dfrac{AB}{\sqrt{3}}=\sqrt{3}AB-150$
Now multiplying the whole equation by $\sqrt{3}$ we get.
$\begin{align}
  & AB=3AB-150\sqrt{3} \\
 & \Rightarrow 2AB=150\sqrt{3} \\
 & \Rightarrow AB=75\sqrt{3} \\
\end{align}$
Now value of $\sqrt{3}$ is given as 1.73
Hence $AB=75\times 1.73=129.75$

Hence the height of the tower is 129.75m

Note: Note that here we want to consider the opposite and adjacent side of the angle hence we have taken tan of the angle. Similarly we can also take cot angle and solve but sin and cos won’t work.