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Text Solution

2520`10sqrt(5)``(100)/(3sqrt(3))`

Answer :

BSolution :

Given ABC is a triangular park with AB = AC = 100 m. A vertical tower is situated at the mid-point of BC. <br> Let the height of the tower is h m. <br> Now, according to given information, we have the following figure. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/41Y_SP_MATH_C20_E04_003_S01.png" width="80%"> <br> From the figure and given information, we have <br> `beta=cot^(-1)(3sqrt(2))` <br> and `alpha=cosec^(-1)(2sqrt(2))` <br> Now, in `DeltaQPA`, <br> `cotbeta = (l)/(h)` <br> `implies l=(3sqrt(2))h" ...(i)"` <br> and in `DeltaBPQ, tan alpha = (h)/(BP)` <br> `implies cot alpha=(BP)/(h)=(sqrt((100^(2))-l^(2)))/(h)` <br> `[because " p is mid-point of isosceles "DeltaABC, AP _|_ BC]` <br> `implies h^(2)cot^(2)alpha=(100)^(2)-l^(2)` <br> `impliesh^(2)(cosec^(2)alpha-1)=(100)^(2)-(3sqrt(2)h)^(2)" [from Eq. (i)]"` <br> `implies h^(2)(8-1)=(100)^(2)-18h^(2)` <br> `implies 25h^(2)=(100)^(2)` <br> `implies h^(2)=((100)/(5))^(2)implies h = 20 m`