Question

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 pole is 24 m, find the height of the first pole. $\left( \sqrt{3}=1.732 \right)$

Answer 1

Hint: Assume the height of the first pole be ‘h’. Draw a rough diagram of the given conditions and then use the formula $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$ in the different right angle triangles and substitute the given values to get the height.


Complete step-by-step answer:


Let us start with the question by drawing a representative diagram of the situation given in the question.

According to the above figure:
BD is the second pole and we have assumed its height to be ‘h’ meters. AE is the second pole with height 24 meters as given in the question.
Now, in right angle triangle ABM,
$\angle ABM=30{}^\circ $ , as it is equal to the angle of depression mentioned in the question and shown in the diagram according to the property of alternate interior angles.
We know that, $\tan \theta =\dfrac{\text{perpendicular}}{\text{base}}$. Therefore,
$\tan {{30}^{\circ }}=\dfrac{AM}{BM}$
Since, BM = DE = 15 m, because they are opposite sides of the rectangle BDEM. Therefore,
$\tan {{30}^{\circ }}=\dfrac{AM}{15}$
Substituting $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$, we get,
$AM=\dfrac{15}{\sqrt{3}}=5\sqrt{3}$
Now from the figure, we can say that $BD=24-AM=24-5\sqrt{3}$
Now we will put the value of root 3 as given in the question.
$\therefore BD=24-5\sqrt{3}=24-5\times 1.732=24-8.66=15.34m$
So, the height of the first pole is 15.34 m.

Note: Be careful about the trigonometric ratios which you are using for the question. We have used the tangent of the given angle because we have to find the height and we have a common base in two right angle triangles that can be easily cancelled.