Question

A tower is \[51\]m high and has a mark at a height of \[25\] m from the ground. Find at what distance the two parts subtend equal angles to an eye at the height of \[5\]m, from the ground.

Answer 1

Hint: At first, we will consider the diagram given below. Then by using Pythagoras theorem we can find the solution of the given problem.

Complete step-by-step answer:
               

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.

Note: Pythagoras theorem states that, for a right-angle triangle, the square of the hypotenuse is equal to the sum of the square of base and the square of perpendicular.
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.