Question

On a horizontal plane there is vertical tower with a flag pole on the top of the tower. At a point 9 metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are 60° and 30° respectively. Find the height of the tower and the flag pole mounted on it.

Answer 1

Answer:

The height of the tower is 3√3 m  & the height of the flag mounted on it is 6√3 m .

Step-by-step explanation:

Let the height of the flag-pole (AD) be  x &  height of tower (AB) be h .

Angle of elevation 9 m away from the foot of the tower to the  top of the  flag-pole , ∠BCD = 60° & angle of elevation to the  bottom of the flag-pole , ∠ACB = 30°

Let BC = 9 m  

In right triangle , ∆ABC,

tan 30° =  P/B = AB/BC  

1/√3 = h/9

√3 h  = 9  

h = 9/√3  

h = (9 ×√3) /(√3 × √3)

[On Rationalising]

h = 9√3/3

h = 3√3

Height of the tower ,h = 3√3 m   ……….(1)

In right triangle , ∆DBC,

tan 60° = P/H = DB/BC  

tan 60° = (AD + AB)/BC

√3 = (h + x)/9

9√3 = (h + x)

9√3 = 3√3 + x

[From eq 1]

9√3 - 3√3 = x

x = 6√3  

Height of the flag mounted on it = 6√3 m  

Hence, the height of the tower is 3√3 m  & the height of the flag mounted on it is 6√3 m .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

SOLUTION

Let the height of the Flag-pole= h(m)

& height of tower = x(m)

In ∆ABC,

$tan60 \degree = \frac{</strong><strong>AB</strong><strong>}{</strong><strong>BC</strong><strong>} \\ \\ = > \sqrt{3} = \frac{h + x}{9} \\ \\ = > h + x = 9 \sqrt{3} .............(1)$

In ∆DBC,

$tan30 \degree = \frac{</strong><strong>DB</strong><strong>}{</strong><strong>BC</strong><strong>} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{x}{9} \\ \\ = > \sqrt{3x} =9 \\ \\ = > x = \frac{9}{ \sqrt{3} } = > \frac{9 \sqrt{3} }{3} \\ \\ = > x = 3 \sqrt{3} .............(2)$

Now, substituting value of x in eqn.(1)

=) h+3√3= 9√3

=) h=9√3-3√3

=) h= 6√3m.

Therefore,

Height of tower is 3√3m& Height of Flag-pole is 6√3m.

hope it helps ☺️