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(B) Let $OP$ be the original pole,which breaks at point $A$. Let the top $P$ strike the ground at $P^{\prime}$.Given: $OP^{\prime} = 21 \ m$ and $\ang

Vedclass · Accepted Answer

$21 \sqrt{3}$

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(B) Let $OP$ be the original pole,which breaks at point $A$. Let the top $P$ strike the ground at $P^{\prime}$.Given: $OP^{\prime} = 21 \ m$ and $\angle OP^{\prime}A = 30^{\circ}$.In $	riangle AOP^{\prime}$,$	an 30^{\circ} = \frac{OA}{OP^{\prime}}$$\frac{1}{\sqrt{3}} = \frac{OA}{21} \Rightarrow OA = \frac{21}{\sqrt{3}} = 7\sqrt{3} \ m$.Also,$\cos 30^{\circ} = \frac{OP^{\prime}}{AP^{\prime}}$$\frac{\sqrt{3}}{2} = \frac{21}{AP^{\prime}} \Rightarrow AP^{\prime} = \frac{42}{\sqrt{3}} = 14\sqrt{3} \ m$.Since the broken part $AP$ is equal to $AP^{\prime}$,the total height of the pole is $OA + AP = OA + AP^{\prime}$.Total height $= 7\sqrt{3} + 14\sqrt{3} = 21\sqrt{3} \ m$.

$A$ pole is broken by the wind. The top of the pole strikes the ground at an angle of $30^{\circ}$ and at a distance of $21 \ m$ from the foot of the pole. Find the total height of the pole (in $m$).

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