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

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(B) Let $BC$ be the height of the tower and $CD$ be the height of the flagstaff, where $BC = x$ and $CD = h$.Let the point be $A$ at a distance $AB =

Vedclass · Accepted Answer

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

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(B) Let $BC$ be the height of the tower and $CD$ be the height of the flagstaff, where $BC = x$ and $CD = h$.Let the point be $A$ at a distance $AB = y$ from the foot of the tower $B$.Given that the tower and the flagstaff subtend equal angles $	heta$ at point $A$.In $	riangle ABC$, $	an 	heta = \frac{BC}{AB} = \frac{x}{y}$.In $	riangle ABD$, the total angle is $2	heta$ and the total height is $BD = BC + CD = x + h$.Thus, $	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}$$2xy^2 = (x+h)(y^2-x^2)$$\frac{2xy^2}{y^2-x^2} = x+h$$h = \frac{2xy^2}{y^2-x^2} - x$$h = \frac{2xy^2 - x(y^2-x^2)}{y^2-x^2}$$h = \frac{2xy^2 - xy^2 + x^3}{y^2-x^2}$$h = \frac{xy^2 + x^3}{y^2-x^2} = \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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