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It is given that the height of the tower is \[51\]m. It has a mark at a height of \[25\]m from the ground. We have to find the distance where the two parts subtend equal angles to an eye at the height of \[5\]m, from the ground.

Let us consider the diagram where,\[AC = 51\], \[B\] is the marked point such that, \[AB = 25\]. Again, \[PQ = AD = 5\]

It is given that,

\[\angle CQB = \angle AQB\].

Therefore \[QB\] is bisector of angle \[\angle AQC\] and as such it divides the base \[AC\] in the ratio of the arm of the angle

\[\dfrac{{AB}}{{BC}} = \dfrac{{QA}}{{QC}}\]

We are going to use Pythagoras theorem for following triangles,

From \[\Delta QDA\] we have, \[\sqrt {Q{D^2} + A{D^2}} = QA\]

From \[\Delta QDC\] we have, \[\sqrt {Q{D^2} + D{C^2}} = QC\]

Let us substitute these values of the sides in the above equation we get,

\[\dfrac{{AB}}{{BC}} = \dfrac{{\sqrt {Q{D^2} + D{A^2}} }}{{\sqrt {Q{D^2} + D{C^2}} }}\]

Let us consider, \[PA = QD = x\]

Let substitute these known as values we get,

\[\dfrac{{25}}{{26}} = \dfrac{{\sqrt {{x^2} + {5^2}} }}{{\sqrt {{x^2} + {{46}^2}} }}\]

On squaring both sides and cross multiplying we get,

\[625({x^2} + {46^2}) = 676({x^2} + 25)\]

Om simplifying the above equation, we get,

\[51{x^2} = (625 \times {46^2} - 676 \times 25)\]

Simplifying again we get,

\[{x^2} = 25600\]

Taking square root on both sides of the above equation we get,

\[x = 160\]

Hence, the required distance is \[160\]m.

To find the value of x diagram plays a major role, it will help us by leading it to find the triangle to which the Pythagoras theorem is applied.