A man from the top of a 100 m high tower see | vedclass

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(A) Let the height of the tower be $h = 100 \ m$.Let $x$ be the distance travelled by the car and $y$ be the distance from the base of the tower to th

Vedclass · Accepted Answer

$\frac{200\sqrt{3}}{3} \ m$

Vedclass · Answer

(A) Let the height of the tower be $h = 100 \ m$.Let $x$ be the distance travelled by the car and $y$ be the distance from the base of the tower to the car's final position.In the right-angled triangle with angle $60^{\circ}$:$	an 60^{\circ} = \frac{100}{y}$$\sqrt{3} = \frac{100}{y} \implies y = \frac{100}{\sqrt{3}} \ m$  ........$(i)$In the right-angled triangle with angle $30^{\circ}$:$	an 30^{\circ} = \frac{100}{x+y}$$\frac{1}{\sqrt{3}} = \frac{100}{x+y} \implies x+y = 100\sqrt{3} \ m$  .....$(ii)$Substituting $(i)$ into $(ii)$:$x = 100\sqrt{3} - \frac{100}{\sqrt{3}}$$x = \frac{100(3) - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \ m$

$A$ man from the top of a $100 \ m$ high tower sees a car moving towards the tower. The angle of depression is $30^{\circ}$. After some time,the angle of depression becomes $60^{\circ}$. The distance travelled by the car during this time is:

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