Question

the angle of depression from the top of the vertical tower to a point on the ground is found to be 60 degree and from a point 50m above the foot of the tower the angle of depression to the same point is found to be 30 degree as shown in the figure find the height of the tower
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Answer 1

Required diagram

Answer:

100cm

Step-by-step explanation:

AD=?

tan d = 50/AD

1/

$\frac{1}{ \sqrt{3} } = \frac{50}{x}$

AD

$= 50 \sqrt{3}$

height of the tower

tan D = 50+x/50root 3

tan 60=50+x/50root 3

$\sqrt{3} = \frac{50 + x}{50 \sqrt{3} } \\ 5 + x = 50 \sqrt{3} \times \sqrt{3} \\ \\ 50 + \times = 50 \times 3 \\ x = 150 - 50 \\ x = 100$

Answer 2

Concept introduction:

The angle of depression is a type of angle which is the angle between the line of our eyes or observation and the line of the horizontal plane.

Given:

• Here it is given that the angle of depression from the top of the vertical tower to a point on the ground is found to be $60$°.
• and from a point $50m$ above the foot of the tower the angle of depression to the same point is found to be $30$°.

To find:

We have to find the height of the tower.

Solution:

According to the problem,

Let $x$ be the height of the tower.

$tan30=\frac{50}{x}$

$- > x=50\sqrt{3}$

$- > tan60=\frac{50+x}{50\sqrt{3} } \\- > x=100$

So, the height of the tower is $100m$.