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(1) Trigonometric ratio

(2) Property of similar triangles.

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Considering the height of the tower as AB.

Using Trigonometric ratios:

We know the trigonometric ratio \[\tan \theta \] is given as \[\tan \theta =\dfrac{P}{B}\], where P is the perpendicular of the triangle and B is the base.

In \[\Delta CDE\]

\[\tan \theta =\dfrac{DE}{DC}\]

\[\tan \theta =\dfrac{1.6}{4.8}\]

\[\tan \theta =\dfrac{1}{3}\]

In \[\Delta CBA\]

\[\tan \theta =\dfrac{AB}{BC}\]

We know \[\tan \theta =\dfrac{1}{3}\], substituting the value of \[\tan \theta \] in the expression above

\[\dfrac{1}{3}=\dfrac{AB}{3.2+4.8}\]

Simplifying the expression above we get,

\[\dfrac{1}{3}=\dfrac{AB}{8}\]

By cross multiplying the terms we get the value of AB as,

\[AB=\dfrac{8}{3}\]

Dividing 8 by 3 to get the value as a decimal,

The height of the pole is 2.67m.

Using property of similar triangles:

In \[\Delta CDE\]and \[\Delta CBA\]

We know the both the angles \[\angle CDE\] and \[\angle CBA\] are of \[90{}^\circ \] measure therefore,

\[\angle CDE=\angle CBA\]

We know the both the angles \[\angle DCE\] and \[\angle BCA\] are common in both the triangles,

\[\angle DCE=\angle BCA\]

By (angle-angle) AA similarity criteria

\[\Delta CDE\sim \Delta CBA\]

Now we know the ratio of sides of two similar triangles is equal,

Expressing the above statement mathematically we get,

\[\dfrac{DE}{AB}=\dfrac{CD}{BC}\]

Substituting the values of DE, CD, BC we get,

\[\dfrac{1.6}{AB}=\dfrac{4.8}{8}\]

Cross multiplying the terms we get AB as,

\[AB=\dfrac{1.6\times 8}{4.8}\]

\[AB=\dfrac{8}{3}\]

AB = 2.67m

Hence found by using both trigonometric ratio and property of similar triangles.