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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

The shadow of a tower standing on a level ground is found to be $60 \ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is

From a point on the level ground,the angle of elevation of the top of a pole is $30^{\circ}$. On moving $20 \ m$ nearer to the pole,the angle of elevation becomes $45^{\circ}$. The height of the pole (in metres) is:

The angle of elevation of a tower from a point $A$ due south of it is $30^\circ$ and from a point $B$ due west of it is $45^\circ$. If the height of the tower is $100 \ m$,then $AB = \dots \ m$.

The upper part of a tree broken by the wind makes an angle of $30^\circ$ with the ground. The distance from the root to the point where the top of the tree touches the ground is $10 \, m$. What was the original height of the tree in meters?

Vedclass Test Series

Exam Paper Generator

Online Exam Module