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(B) Let the height of the tower be $CB = 100 \, m$. Let the man walk a distance $AD = x \, m$ towards the tower.In $\Delta DBC$ (right-angled at $B$):

Vedclass · Accepted Answer

$73.2$

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(B) Let the height of the tower be $CB = 100 \, m$. Let the man walk a distance $AD = x \, m$ towards the tower.In $\Delta DBC$ (right-angled at $B$):$	an 45^{\circ} = \frac{CB}{DB}$$1 = \frac{100}{DB}$$DB = 100 \, m$In $\Delta ABC$ (right-angled at $B$):$	an 30^{\circ} = \frac{CB}{AB} = \frac{CB}{AD + DB}$$\frac{1}{\sqrt{3}} = \frac{100}{x + 100}$$x + 100 = 100\sqrt{3}$$x = 100(\sqrt{3} - 1)$Using $\sqrt{3} \approx 1.732$:$x = 100(1.732 - 1) = 100(0.732) = 73.2 \, m$.Thus,the distance walked by the man is $73.2 \, m$.

$A$ man observes the angle of elevation of the top of a tower having a height of $100 \, m$ as $30^{\circ}$. After walking a certain distance towards the tower,he observes the angle of elevation as $45^{\circ}$. What is the distance through which the man walked towards the tower? (in $m$)

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