From a point on the level ground,the angle of | vedclass

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(B) Let the height of the pole be $h$ and the distance from the second point to the base of the pole be $x$. In $	riangle BDA$,$	an 45^{\circ} = \fr

Vedclass · Accepted Answer

$10(\sqrt{3}+1)$

Vedclass · Answer

(B) Let the height of the pole be $h$ and the distance from the second point to the base of the pole be $x$. In $	riangle BDA$,$	an 45^{\circ} = \frac{h}{x}$ $\Rightarrow 1 = \frac{h}{x}$ $\Rightarrow h = x$. In $	riangle BCA$,$	an 30^{\circ} = \frac{h}{20+x}$. Substituting $x = h$,we get $\frac{1}{\sqrt{3}} = \frac{h}{20+h}$. $20 + h = h\sqrt{3} \Rightarrow h(\sqrt{3}-1) = 20$. $h = \frac{20}{\sqrt{3}-1} = \frac{20(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{20(\sqrt{3}+1)}{3-1} = \frac{20(\sqrt{3}+1)}{2} = 10(\sqrt{3}+1) \ m$.

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:

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