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(A) Let $AB$ be the tower of height $h \; m$.Let $C$ be the initial point such that $BC = 160 \; m$ and $\angle ACB = 	heta$.After moving $100 \; m$

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) Let $AB$ be the tower of height $h \; m$.Let $C$ be the initial point such that $BC = 160 \; m$ and $\angle ACB = 	heta$.After moving $100 \; m$ towards the tower,we reach point $D$,such that $CD = 100 \; m$. Thus,$BD = BC - CD = 160 - 100 = 60 \; m$.At point $D$,the angle subtended is $2	heta$,so $\angle ADB = 2	heta$.In $	riangle ABC$,$	an 	heta = \frac{AB}{BC} = \frac{h}{160}$.In $	riangle ABD$,$	an 2	heta = \frac{AB}{BD} = \frac{h}{60}$.Using the formula $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,we have:$\frac{h}{60} = \frac{2(h/160)}{1 - (h/160)^2}$.$\frac{h}{60} = \frac{h/80}{1 - h^2/25600}$.Dividing by $h$ (since $h 
eq 0$):$\frac{1}{60} = \frac{1/80}{(25600 - h^2)/25600} = \frac{25600}{80(25600 - h^2)} = \frac{320}{25600 - h^2}$.$25600 - h^2 = 320 	imes 60 = 19200$.$h^2 = 25600 - 19200 = 6400$.$h = \sqrt{6400} = 80 \; m$.

$A$ tower standing on a horizontal plane subtends a certain angle at a point $160 \; m$ apart from the foot of the tower. On advancing $100 \; m$ towards it,the tower is found to subtend an angle twice as before. The height of the tower is (in $m$):

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