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.

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Vedantu Content Team · Accepted Answer

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:               \n    \n    \n    \n    \n  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,$${x^2} = 25600$$\t\t\t\t\t\t\t\t\t\t\tTaking 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.

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.