The tops of two poles of height 60 m and 35 | vedclass

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Question

(D) Let the two poles be $AB$ and $CD$ with heights $60 \ m$ and $35 \ m$ respectively.Let $DE$ be the horizontal line from $D$ to the pole $AB$,meeti

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Let the two poles be $AB$ and $CD$ with heights $60 \ m$ and $35 \ m$ respectively.Let $DE$ be the horizontal line from $D$ to the pole $AB$,meeting $AB$ at $E$.Then $BE = CD = 35 \ m$.The height $AE = AB - BE = 60 - 35 = 25 \ m$.In the right-angled triangle $	riangle AED$,the angle of depression/elevation is $\phi$.Given $	an \phi = \frac{5}{9}$.In $	riangle AED$,$	an \phi = \frac{AE}{DE}$.Substituting the values,$\frac{5}{9} = \frac{25}{DE}$.$DE = \frac{25 	imes 9}{5} = 5 	imes 9 = 45 \ m$.Thus,the distance between the two poles is $45 \ m$.

The tops of two poles of height $60 \ m$ and $35 \ m$ are connected by a rope. If the rope makes an angle with the horizontal whose tangent is $\frac{5}{9}$,then what is the distance (in $m$) between the two poles?

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