The angle of elevation of the top of a tower is observed to be $ 60^{\circ} . $ At a point, $ 30 \mathrm{m} $ vertically above the first point of observation, the elevation is found to be $ 45^{\circ} . $ Find :

(i) the height of the tower,

(ii) its horizontal distance from the points of observation.

In the above diagram AB represents the tower. Point C is the first point of observation and angle of elevation from that point is 60^o. Point D is 30 m above point C and angle of elevation from that point is 45^o. ED is parallel to BC. 

Now,

Let BC = x meter and AE = h meter.

Now,

In ∆ABC:

$ \begin{array}{l}
tan\ 60\ =\ \frac{h\ +\ 30}{x}\\
\\
\\
\sqrt{3} \ =\ \frac{h\ +\ 30}{x}\\
\\
\\
x\sqrt{3} \ =\ h\ +\ 30\ \ \ \ \ \ ...( i)
\end{array}$

In ∆ADE:

$ \begin{array}{l}
tan\ 45\ =\ \frac{h}{x}\\
\\
\\
1\ =\ \frac{h}{x}\\
\\
\\
x\ =\ h\ \ \ \ \ \ \ \ ...( ii)
\end{array}$

Putting value of x in eq (i):

$ \begin{array}{l}
h\ =\ \frac{30}{\sqrt{3} \ -\ 1}\\
\\
\\
h\ =\ \frac{30}{1.732\ -\ 1}\\
\\
\\
h\ =\ \frac{30}{0.732}\\
\\
\\
h\ =\ 40.98\ m\\
\\
\\
h\ =\ 41\ m
\end{array}$

So, from eq (ii):

x = 41 m

Now,

i)  Height of the tower = h + 30 = 41 + 30 = 71 m

ii)  Horizontal distance from the point of observation = x = 41 m

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Updated on: 10-Oct-2022

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