(B) Let the height of the tower be $PQ = 100 \ m$. Let the initial position of the car be $R$ and the final position be $S$. The distance travelled by

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(B) Let the height of the tower be $PQ = 100 \ m$. Let the initial position of the car be $R$ and the final position be $S$. The distance travelled by the car is $x = RS$.In $\triangle PQS$,$\tan{60^\circ} = \frac{PQ}{QS} \implies \sqrt{3} = \frac{100}{QS} \implies QS = \frac{100}{\sqrt{3}} \ m$.In $\triangle PQR$,$\tan{30^\circ} = \frac{PQ}{QR} \implies \frac{1}{\sqrt{3}} = \frac{100}{QR} \implies QR = 100\sqrt{3} \ m$.The distance travelled by the car is $x = QR - QS = 100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

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$A$ man from the top of a $100 \ m$ high tower sees a car moving towards the tower at an angle of depression of $30^\circ$. After some time,the angle of depression becomes $60^\circ$. The distance (in meters) travelled by the car during the time is

IIT 2001 All IIT

A
$100\sqrt{3}$
B
$\frac{200\sqrt{3}}{3}$
C
$\frac{100\sqrt{3}}{3}$
D
$200\sqrt{3}$

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$A$ man from the top of a $100 \ m$ high tower sees a car moving towards the tower at an angle of depression of $30^\circ$. After some time,the angle of depression becomes $60^\circ$. The distance (in meters) travelled by the car during the time is

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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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