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(A) Let the height of the tower $AB = 1 \ m$ and the height of the flag $BC = x \ m$. Let the point be $D$ at a distance $AD = 2 \ m$ from the foot of

Vedclass · Accepted Answer

$\frac{5}{3}$

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(A) Let the height of the tower $AB = 1 \ m$ and the height of the flag $BC = x \ m$. Let the point be $D$ at a distance $AD = 2 \ m$ from the foot of the tower.Let $\angle BDA = \alpha$ and $\angle CDA = 2\alpha$ (since the flag and tower subtend equal angles $\alpha$ at $D$,the total angle is $2\alpha$).In $	riangle ABD$,$	an \alpha = \frac{AB}{AD} = \frac{1}{2}$.In $	riangle ACD$,$	an 2\alpha = \frac{AC}{AD} = \frac{1+x}{2}$.Using the formula $	an 2\alpha = \frac{2 	an \alpha}{1 - 	an^2 \alpha}$,we get:$\frac{1+x}{2} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.Therefore,$\frac{1+x}{2} = \frac{4}{3}$ $\Rightarrow 1+x = \frac{8}{3}$ $\Rightarrow x = \frac{8}{3} - 1 = \frac{5}{3} \ m$.

$A$ flag is standing vertically on a tower of height $1 \ m$. At a point at a distance $2 \ m$ from the foot of the tower,the flag and the tower subtend equal angles. The height of the flag is (in meters):

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