ABC is a triangular park with AB = AC = 100 t | vedclass

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(D) Let $P$ be the mid-point of $BC$. In $	riangle ABC$,$AB=AC=100$. Let $AP = x$. Since $P$ is the mid-point of $BC$,$AP \perp BC$.In $	riangle ABP

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) Let $P$ be the mid-point of $BC$. In $	riangle ABC$,$AB=AC=100$. Let $AP = x$. Since $P$ is the mid-point of $BC$,$AP \perp BC$.In $	riangle ABP$,$BP^2 = AB^2 - AP^2 = 100^2 - x^2$. Thus $BP = \sqrt{10000 - x^2}$.Let the height of the tower $PQ = h$.Given the angle of elevation at $A$ is $\alpha = \cot^{-1}(3\sqrt{2})$,so $\cot \alpha = \frac{AP}{PQ} = \frac{x}{h} = 3\sqrt{2} \implies x = 3\sqrt{2}h$.Given the angle of elevation at $B$ is $\beta = \csc^{-1}(2\sqrt{2})$,so $\csc \beta = \frac{BQ}{PQ} = \frac{\sqrt{BP^2 + h^2}}{h} = 2\sqrt{2}$.Squaring both sides: $\frac{BP^2 + h^2}{h^2} = 8 \implies BP^2 + h^2 = 8h^2 \implies BP^2 = 7h^2$.Substitute $BP^2 = 100^2 - x^2$: $10000 - x^2 = 7h^2$.Substitute $x = 3\sqrt{2}h$: $10000 - (3\sqrt{2}h)^2 = 7h^2 \implies 10000 - 18h^2 = 7h^2$.$25h^2 = 10000 \implies h^2 = 400 \implies h = 20$.

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