The horizontal distance between two poles is $15 m .$ The angle of depression of the top of the first pole as seen from the top of the second pole is $30^{\circ}$ . If the height of the second pole is $24 m .$ Find the height of the first pole.

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Solution

Let the length of the side $B E$ be $x .$

Now, in $△ B E D,$
$\begin{matrix} tan 30^{\circ} & = \frac{B E}{E D} \frac{1}{\sqrt{3}} & = \frac{x}{15} x & = 5 \sqrt{3} m \end{matrix}$

So, height of first pole will be,
$A B = A E + B E$
$= 24 m + 5 \sqrt{3} m$
$= (24 + 5 \sqrt{3}) m$

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The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30∘. If the height of the second pole is 24 m. Find the height of the first pole.

Let the length of the side BE be x.Now, in △BED,tan30∘=BEED1√3=x15x=5√3 mSo, height of first pole will be,AB=AE+BE =24m+5√3m =(24+5√3)m

The horizontal distance between two poles is $15 m .$ The angle of depression of the top of the first pole as seen from the top of the second pole is $30^{\circ}$ . If the height of the second pole is $24 m .$ Find the height of the first pole.

Let the length of the side $B E$ be $x .$

Now, in $△ B E D,$
$\begin{matrix} tan 30^{\circ} & = \frac{B E}{E D} \frac{1}{\sqrt{3}} & = \frac{x}{15} x & = 5 \sqrt{3} m \end{matrix}$

So, height of first pole will be,
$A B = A E + B E$
$= 24 m + 5 \sqrt{3} m$
$= (24 + 5 \sqrt{3}) m$