Question

The angle of depression from the top of a tower of a point A on the ground is 30 degrees.On moving distance of 20m from the point A towards the foot of the tower to a point B ,the angle of elevation of the top of the tower from the point B is 60 degrees.Find the height of the tower and its distance from point A

Answer 1

Answer:

The height of tower is 10√3 m

Distance of tower from point A is 30 m

Step-by-step explanation:

Refer the attached figure

Height of tower = CD =x

The angle of depression from the top of a tower of a point A on the ground is 30 degrees. i.e. ∠DAC = 30°

Moving distance of 20m from the point A towards the foot of the tower to a point B i.e. AB = 20 m

The angle of elevation of the top of the tower from the point B is 60 degrees. i.e. ∠DBC= 60°

Let CB be x

AC = CB +BA = x+20

Now in ΔDCB

Using trigonometric ratios

$tan\theta = \frac{Perpendicular}{Base}$

$tan 60 ^{\circ} = \frac{DC}{CB}$

$\sqrt{3} = \frac{h}{x}$

$\sqrt{3}x = h$     ----1

Now in ΔDCA

$tan\theta = \frac{Perpendicular}{Base}$

$tan 30 ^{\circ} = \frac{DC}{CA}$

$\frac{1}{\sqrt{3}}= \frac{h}{x+20}$

$\frac{x+20}{\sqrt{3}} = h$   --2

Equate 1 and 2

$\frac{x+20}{\sqrt{3}} = \sqrt{3}x$

$x+20= 3x$

$20= 2x$

$\frac{20}{2}=x$

$10=x$

Now Substitute the value in 1

$10\sqrt{3} m= h$

Distance of tower from point A = CA = x+20=10+20=30 m

Hence the height of tower is 10√3 m

Distance of tower from point A is 30 m

Answer 2

hope its help you thanks