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\[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\]

\[\tan {60^ \circ } = \sqrt 3 \]

It is given that; a man on the top of a bamboo pole observes that the angle of depression of the base and the top of another pole is \[{60^ \circ }\,\&\, {30^ \circ }\] respectively. The length of the second pole is \[5\]m above the ground level.

We have to find the height of the bamboo pole on which the man is sitting.

As per the given problem, \[BD = 5\] and the angle of depression is given by \[\angle ABC = {30^ \circ }\,\&\, \angle ADE = {60^ \circ }\]

The man is sitting at the point A.

Let us consider, the length of \[AC\] be \[x\].

Since, \[BCED\] is a rectangle, the opposite sides are equal.

So, we have, \[BD = CE = 5\]

So, the length of the bamboo on which the man is sitting is \[AE = AC + CE\]

Now we find the length of \[AC.\]

From, \[\Delta ABC,\] we get,

\[ \Rightarrow \dfrac{{AC}}{{BC}} = \tan {30^ \circ }\]

Substitute the values we get,

\[ \Rightarrow \dfrac{x}{{BC}} = \dfrac{1}{{\sqrt 3 }}\]

Simplifying we get,

\[\therefore BC = x\sqrt 3 \]

From, \[\Delta ADE,\] we get,

\[ \Rightarrow \dfrac{{AE}}{{ED}} = \tan {60^ \circ }\]

Substitute the values we get,

\[ \Rightarrow \dfrac{{x + 5}}{{DE}} = \sqrt 3 \]

Simplifying we get,

\[\therefore DE = \dfrac{{x + 5}}{{\sqrt 3 }}\]

We already know that, \[DE = BC\]

Equating the length, we get,

\[ \Rightarrow x\sqrt 3 = \dfrac{{x + 5}}{{\sqrt 3 }}\]

Simplifying we get,

\[ \Rightarrow 3x = x + 5\]

Simplifying, again we get,

\[ \Rightarrow 2x = 5\]

\[\therefore x = \dfrac{5}{2} = 2.5\]

So, the length of the bamboo is \[AE = 5 + 2.5 = 7.5\]cm

Hence, the length of the bamboo is \[7.5\] cm.