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

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(B) Let $H$ be the height of the tower $AB$ and $B$ be the base of the tower.Let $C$ be the initial position and $D$ be the position after walking $10

Vedclass · Accepted Answer

$50 \sqrt{3}$

Vedclass · Answer

(B) Let $H$ be the height of the tower $AB$ and $B$ be the base of the tower.Let $C$ be the initial position and $D$ be the position after walking $100 \ m$ towards the tower.Thus,$CD = 100 \ m$.In $\Delta ABD$,$	an 60^{\circ} = \frac{AB}{DB} = \frac{H}{DB}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $DB = \frac{H}{\sqrt{3}}$ ... $(1)$.In $\Delta ABC$,$	an 30^{\circ} = \frac{AB}{BC} = \frac{H}{CD + DB} = \frac{H}{100 + DB}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{H}{100 + DB}$,which implies $100 + DB = H\sqrt{3}$,or $DB = H\sqrt{3} - 100$ ... $(2)$.Equating $(1)$ and $(2)$,we get $\frac{H}{\sqrt{3}} = H\sqrt{3} - 100$.Multiplying by $\sqrt{3}$,we get $H = 3H - 100\sqrt{3}$.$2H = 100\sqrt{3}$,so $H = 50\sqrt{3} \ m$.

The angle of elevation of the top of a tower is $30^{\circ}$. On walking $100 \ m$ towards the tower,the angle of elevation of the top of the tower becomes $60^{\circ}$. Find the height of the tower (in $m$).

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