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We have given that a pole 5 m high is fixed on the top of a tower. The angle of elevation of the top the pole observed from a point A on the ground is $ 60{}^\circ $ and the angle of depression of the point A from the top of the tower is $ 45{}^\circ $ .

We have to find the height of the tower.

First we draw a diagram assuming a point of observation A at the ground.

We have given that height of pole is $ 5\text{ m} $ . i.e. $ BC=5m $

Let us assume the height of tower is $ x $ , i.e. $ CD=x $.

Also, we have given angle of elevation $ \angle BAC=60{}^\circ $ and angle of depression $ \angle ACO=45{}^\circ $.

First let us consider right –angled triangle $ \Delta ADB $ ,

We know that $ \tan \theta =\dfrac{\text{Perpendicular}}{\text{Base}} $

We have $ \theta =60{}^\circ $ as given in the question, angle of elevation.

When we substitute the values, we get

$ \begin{align}

& \tan 60{}^\circ =\dfrac{BD}{AD} \\

& \tan 60{}^\circ =\dfrac{x+5}{AD} \\

\end{align} $

We know that $ \tan 60{}^\circ =\sqrt{3} $ , so substitute the value we get

$ \begin{align}

& \sqrt{3}=\dfrac{x+5}{AD} \\

& \sqrt{3}AD=x+5 \\

& AD=\dfrac{x+5}{\sqrt{3}}................(i) \\

\end{align} $

Now, consider right –angled triangle $ \Delta ADC $ ,

We know that $ \tan \theta =\dfrac{\text{Perpendicular}}{\text{Base}} $

Now, from the diagram we have $ \angle OCD=90{}^\circ $

Also,

$ \begin{align}

& \angle OCD=\angle ACD+\angle ACO \\

& 90{}^\circ =\angle ACD+45{}^\circ \\

& \angle ACD=45{}^\circ \\

\end{align} $

So, we have $ \theta =45{}^\circ $

When we substitute the values, we get

$ \begin{align}

& \tan 45{}^\circ =\dfrac{AD}{CD} \\

& \tan 45{}^\circ =\dfrac{AD}{x} \\

\end{align} $

We know that $ \tan 45{}^\circ =1 $

$ \begin{align}

& 1=\dfrac{AD}{x} \\

& AD=x.......(ii) \\

\end{align} $

When we put the value of $ AD $ from equation (i), we get

$ \begin{align}

& \dfrac{x+5}{\sqrt{3}}=x \\

& x+5=\sqrt{3}x \\

\end{align} $

We know that $ \left( \sqrt{3}=1.732 \right) $ ,

$ \begin{align}

& x+5=1.732x \\

& 5=1.732x-x \\

& 5=0.732x \\

& x=\dfrac{5}{0.732} \\

& x=6.83m \\

\end{align} $

Height of the tower is $ 6.83m $.