Question

A pole \[5m\] high is fixed on the top of the tower. The angle of elevation of the top of the poles as observed from a point \[A\] on the ground is \[60\] degree and the angle of depression of the point\[A\] from the top of the tower is \[45\]degree. Find the height of the tower.

Answer 1

Hint: Draw the figure of the tower then using a vertically opposite angle concept, try to equal the angles.Using the trigonometric ratio concept , find the height of the tower by taking two different triangles and solve for the height.

Formula used:
Tangent trigonometric ratio concept.
\[\tan {{60}^{\circ }}=\sqrt{3}\]

Complete step by step answer:
irst, by the help of the above information, we draw the figure of the question.
 Let \[BC\] be the height of the tower and \[CD\] be the pole of height \[5m\] fixed on the top of the tower.
Take
\[BC=hm\].
Also,
The angle of elevation of top of the pole from the point \[A\] on the ground be \[60\] degree and the angle of depression of the point \[A\] from the top of the tower be \[45\] degree, that is \[\angle BAD={{60}^{\circ }}\] and \[\begin{align}
  & \angle BAC={{45}^{\circ }} \\
 & \\
\end{align}\]
 In right angled triangle\[ABC\], we have;
\[\begin{align}
  & \tan {{45}^{\circ }}=\dfrac{BC}{AB} \\
 & \Rightarrow 1=\dfrac{h}{AB} \\
 & \Rightarrow AB=h \\
\end{align}\]
Also in the right angled triangle\[ABD\], we have;
\[\begin{align}
  & \tan {{60}^{\circ }}=\dfrac{BD}{AB} \\
 & \Rightarrow \sqrt{3}=\dfrac{BC+CD}{AB} \\
 & \Rightarrow \sqrt{3}=\dfrac{h+5}{AB} \\
 & \Rightarrow AB=\dfrac{h+5}{\sqrt{3}} \\
\end{align}\]
Comparing both the equation, we get;
\[\begin{align}
  & \Rightarrow \sqrt{3}h-h=5 \\
 & \Rightarrow h(\sqrt{3}-1)=5 \\
\end{align}\]
Now on rationalizing, we obtain;
\[\Rightarrow h=\dfrac{5}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}\]
\[\Rightarrow h=\dfrac{5(\sqrt{3}+1)}{(\sqrt{3{{)}^{2}}}-{{1}^{2}}}=\dfrac{5(\sqrt{3}+1)}{2}\]
\[AB\] is equal to \[\dfrac{h+5}{\sqrt{3}}\] and \[h\] has above equation, so solving these we get;
\[\begin{align}
  & \Rightarrow h=\dfrac{5(\sqrt{3}+1)}{2} \\
 & \Rightarrow h=\dfrac{5\times (1.732+1)}{2} \\
 & \Rightarrow h=\dfrac{5\times (2.732)}{2} \\
 & \Rightarrow h=\dfrac{13.66}{2} \\
 & \Rightarrow h=6.83m \\
 & \\
\end{align}\]
Putting the value of \[\sqrt{3}\] that is equal to \[1.732\] and solving for \[h\] we get;
\[\begin{align}
  & h=6.83m \\
 & \\
\end{align}\]
That is the height of the tower is \[6.83m\].

Note: We have to only find the height of the tower so do not consider the triangle for the poles.
Use an alternative angle concept to make the lower angles equal.