Let a vertical tower AB of height 2h stand on | vedclass

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(C) Let $A$ be the foot of the tower and $B$ be the top. The height of the tower is $AB = 2h$. Let $C$ be a point on the tower such that $AC = h$.From

Vedclass · Accepted Answer

$\sqrt{7}-2$

Vedclass · Answer

(C) Let $A$ be the foot of the tower and $B$ be the top. The height of the tower is $AB = 2h$. Let $C$ be a point on the tower such that $AC = h$.From point $P$ on the ground,the angle of elevation to $C$ is $2\alpha$. Let $AP = x$. In $	riangle PAC$,$	an 2\alpha = \frac{AC}{AP} = \frac{h}{x}$. Thus,$x = h \cot 2\alpha$.When the man moves a distance $d = \sqrt{7}h$ towards the tower to point $H$,the distance $AH = x + \sqrt{7}h$. The angle of elevation to the top $B$ is $\alpha$. In $	riangle HAB$,$	an \alpha = \frac{AB}{AH} = \frac{2h}{x + \sqrt{7}h}$.Substituting $x = h \cot 2\alpha$,we get $	an \alpha = \frac{2h}{h \cot 2\alpha + \sqrt{7}h} = \frac{2}{\cot 2\alpha + \sqrt{7}}$.Since $\cot 2\alpha = \frac{1 - 	an^2 \alpha}{2 	an \alpha}$,we have $	an \alpha = \frac{2}{\frac{1 - 	an^2 \alpha}{2 	an \alpha} + \sqrt{7}}$.Let $t = 	an \alpha$. Then $t = \frac{4t}{1 - t^2 + 2\sqrt{7}t}$.Since $t 
eq 0$,$1 = \frac{4}{1 - t^2 + 2\sqrt{7}t}$,so $1 - t^2 + 2\sqrt{7}t = 4$.$t^2 - 2\sqrt{7}t + 3 = 0$.Using the quadratic formula,$t = \frac{2\sqrt{7} \pm \sqrt{(2\sqrt{7})^2 - 4(1)(3)}}{2} = \frac{2\sqrt{7} \pm \sqrt{28 - 12}}{2} = \frac{2\sqrt{7} \pm 4}{2} = \sqrt{7} \pm 2$.Since $\alpha$ is an angle of elevation,$	an \alpha$ must be positive. Also,for $	an 2\alpha$ to be defined and positive,$2\alpha  1$,we take $	an \alpha = \sqrt{7} - 2$.

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