(D) Let the height of the tower $MN = h$.In $\Delta QMN$,we have $\tan 30^o = \frac{MN}{QM}$.$\therefore QM = \frac{h}{\tan 30^o} = h\sqrt{3}$. Since

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(D) Let the height of the tower $MN = h$.In $\Delta QMN$,we have $\tan 30^o = \frac{MN}{QM}$.$\therefore QM = \frac{h}{\tan 30^o} = h\sqrt{3}$. Since $M$ is the midpoint of $QR$,$QM = MR = h\sqrt{3}$.In $\Delta MNP$,the angle of elevation of the top of the tower at $P$ is $45^o$,so $\tan 45^o = \frac{MN}{PM}$.$\therefore PM = \frac{h}{\tan 45^o} = h$.In $\Delta PMQ$,since $PM \perp MQ$ (as the tower is vertical to the ground plane),we have $PQ^2 = PM^2 + QM^2$.Substituting the values: $(200)^2 = h^2 + (h\sqrt{3})^2$.$40000 = h^2 + 3h^2$.$40000 = 4h^2$.$h^2 = 10000$.$h = 100 \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

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In a triangular park $PQR$,$PQ = PR = 200 \ m$. $A$ $TV$ tower stands at the mid-point $M$ of $QR$. If the angles of elevation of the top of the tower at $P, Q,$ and $R$ are $45^o, 30^o,$ and $30^o$ respectively,then the height of the tower (in $m$) is:

JEE MAIN 2018 All JEE MAIN

A
$50$
B
$100\sqrt{3}$
C
$50\sqrt{2}$
D
$100$

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In a triangular park $PQR$,$PQ = PR = 200 \ m$. $A$ $TV$ tower stands at the mid-point $M$ of $QR$. If the angles of elevation of the top of the tower at $P, Q,$ and $R$ are $45^o, 30^o,$ and $30^o$ respectively,then the height of the tower (in $m$) is:

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