Question

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 50 m 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​

Answer 1

Step-by-step explanation:

Height of the tower, H = (h+10)m

In △ABC,

tan60

∘

=

AB

h+10

AB=

3

h+10

⋯(i)

In △DCO,

tan45

∘

=

AB

h

∴AB=h⋯(ii)

comparing eq.(i) and (ii), we get :

h=

3

h+10

∴

3

h=h+10

⇒

3

h−h=10

⇒h(

3

−1)=10

∴h=

3

−1

10

Height of tower = h + 10

=

3

−1

10

+10

=

3

−1

10+10

3

−10

H=

3

−1

10

3

m

Answer 2

We need to recall the following trigonometric formulas.

• $tan\theta=\frac{Opposite\space\ side}{Adjacent\space\ side}$
• $tan30\textdegree=\frac{1}{\sqrt{3} }$
• $tan60\textdegree=\sqrt{3}$

Given:

The angle of depression from the top of the tower to a point $=60\textdegree$

The angle of depression from a point $50$ m above the foot of the tower to the same point $=30\textdegree$

Let's consider,

$(50+h)$ m be the height of the tower.

$x$ m be the distance between the point and foot of the tower.

From $\triangle ABC$, we get

$tanA=\frac{BC}{AB}$

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

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

$x=50\sqrt{3}$

Thus, the distance between the point and foot of the tower is $50\sqrt{3}$ m.

From $\triangle ABD$, we get

$tanA=\frac{BD}{AB}$

$tan60\textdegree=\frac{50+h}{50\sqrt{3} }$

$\sqrt{3} =\frac{50+h}{50\sqrt{3} }$

$150=50+h$

$h=150-50$

$h=100$

We get,

$50+h=50+100$

$50+h=150$ m

Hence, the height of the tower is $150$ m.