A tower,of x metres high,has a flagstaff at i | vedclass

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(B) Let $BC = x$ be the height of the tower and $CD = h$ be the height of the flagstaff. Let the point be $A$ at a distance $y$ from the foot $B$. In

Vedclass · Accepted Answer

$\frac{x(y^2+x^2)}{(y^2-x^2)}$

Vedclass · Answer

(B) Let $BC = x$ be the height of the tower and $CD = h$ be the height of the flagstaff. Let the point be $A$ at a distance $y$ from the foot $B$. In $	riangle ABC$,$	an 	heta = \frac{BC}{AB} = \frac{x}{y}$. In $	riangle ABD$,the total angle is $2	heta$,so $	an 2	heta = \frac{BD}{AB} = \frac{x+h}{y}$. Using the formula $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$,we have: $\frac{2(x/y)}{1-(x/y)^2} = \frac{x+h}{y}$ $\frac{2x/y}{(y^2-x^2)/y^2} = \frac{x+h}{y}$ $\frac{2xy}{y^2-x^2} = \frac{x+h}{y}$ $2xy^2 = (y^2-x^2)(x+h)$ $h = \frac{2xy^2}{y^2-x^2} - x = \frac{2xy^2 - x(y^2-x^2)}{y^2-x^2} = \frac{2xy^2 - xy^2 + x^3}{y^2-x^2} = \frac{xy^2 + x^3}{y^2-x^2} = \frac{x(x^2+y^2)}{y^2-x^2}$. Note: The option $C$ is $\frac{x(x^2+y^2)}{x^2-y^2}$,which is the negative of our result. Given the standard form of such problems,the correct length is $\frac{x(x^2+y^2)}{y^2-x^2}$.

$A$ tower,of $x$ metres high,has a flagstaff at its top. The tower and the flagstaff subtend equal angles at a point distant $y$ metres from the foot of the tower. Then,the length of the flagstaff (in metres) is:

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