The upper part of a tree broken by the wind m | vedclass

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(A) Let the tree be represented by $PQ$,where $Q$ is the root and $M$ is the point where it breaks. When the top $P$ touches the ground at $A$,we have

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$10 \sqrt{3}$

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(A) Let the tree be represented by $PQ$,where $Q$ is the root and $M$ is the point where it breaks. When the top $P$ touches the ground at $A$,we have a right-angled triangle $	riangle MQA$ with $\angle MAQ = 30^{\circ}$ and $AQ = 10 \ m$.In $	riangle MQA$:$	an 30^{\circ} = \frac{MQ}{AQ} \Rightarrow \frac{1}{\sqrt{3}} = \frac{MQ}{10} \Rightarrow MQ = \frac{10}{\sqrt{3}} \ m$.Also,$\cos 30^{\circ} = \frac{AQ}{AM} \Rightarrow \frac{\sqrt{3}}{2} = \frac{10}{AM} \Rightarrow AM = \frac{20}{\sqrt{3}} \ m$.Since the broken part $MP$ is equal to $AM$,the total height of the tree is $MQ + MP = MQ + AM$.Height $= \frac{10}{\sqrt{3}} + \frac{20}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3} \ m$.

The upper part of a tree broken by the wind makes an angle of $30^{\circ}$ with the ground and the distance from the root to the point where the top of the tree touches the ground is $10 \ m$. What was the height of the tree?

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