Question

Two vertical poles are fixed 60m apart. The angle of depression of the top of the first as seen from the top of the second with their 150 M high is 30 degree. Find the height of the first pole.

Answer 1

Solution: The height of first pole is $170\sqrt{3}$ M.

Explanation:

It is given that the distance between two poles is 60 m and the angle of depression of the top of the first as seen from the top of the second with their 150 M high is 30 degree.

First we draw a figure as shown below,

From the given figure it is noticed that if the angle of depression is 30 degree than the value of angle BAE is 60 degree.

Let the length of AB be x.

In a right angle triangle, We know that $\tan \theta=\frac{perpendicular}{base}$

In triangle ABE,

$\tan (60)=\frac{BE}{AB} \\\sqrt{3} =\frac{60}{x}\\x=\frac{60}{\sqrt{3} }\\x=\frac{60}{\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}} )\\x=\frac{60\sqrt{3}}{3} \\x=20\sqrt{3}$.

The height of first pole is AB+BC = $150+20\sqrt{3} =170\sqrt{3}$.

Therefore, the height of first pole is $170\sqrt{3}$ M.

Answer 2

Answer:

Step-by-step explanation:

Check the attachment