The angle of elevation of the top of a tower | vedclass

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(D) Let $AB = h$ be the height of the tower and $A$ be the foot of the tower.Let $GA = x$ be the initial distance from the tower.In $	riangle GAB$,$\

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$10 \sqrt{3} \ m$

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(D) Let $AB = h$ be the height of the tower and $A$ be the foot of the tower.Let $GA = x$ be the initial distance from the tower.In $	riangle GAB$,$	an 30^{\circ} = \frac{AB}{GA} = \frac{h}{x}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{h}{x} = \frac{1}{\sqrt{3}}$,so $x = h\sqrt{3}$.After walking $20 \ m$ towards the tower,the new distance is $HA = x - 20$.In $	riangle HAB$,$	an 60^{\circ} = \frac{AB}{HA} = \frac{h}{x - 20}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $\frac{h}{x - 20} = \sqrt{3}$,so $h = \sqrt{3}(x - 20)$.Substituting $x = h\sqrt{3}$ into the equation: $h = \sqrt{3}(h\sqrt{3} - 20)$.$h = 3h - 20\sqrt{3}$.$2h = 20\sqrt{3}$.$h = 10\sqrt{3} \ m$.

The angle of elevation of the top of a tower at a point $G$ on the ground is $30^{\circ}$. On walking $20 \ m$ towards the tower,the angle of elevation becomes $60^{\circ}$. The height of the tower is equal to

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