Question

2. From the terrace of a house 8 m high, the angle of elevation
of the top of a tower is 45° and the angle of depression of
its reflection in a lake is found to be 60º. Determine the
height of the tower from the ground.​

Answer 1

Given:

From the terrace of a house 8 m high, the angle of elevation  of the top of a tower are 45° and the angle of depression of  its reflection in a lake is found to be 60º

To find:

The  height of the tower from the ground

Solution:

To solve the above-given problem we will use the following trigonometric ratio of a triangle:

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

Referring to the figure attached below we get,

"AB" = the height of the house = 8 m

"BC" = the distance of the base of the tower and the house

"EC" = the height of the tower

"∠DAC" = "∠ACB" = the angle of depression its reflection in a lake = 60°

"∠EAD" = the angle of elevation  of the top of a tower = 45°

AB = CD = 8 m ..... (i) ....... [since ABCD is a rectangle]

Considering Δ ABC, we have

AB = perpendicular

BC = base

θ = 60°

∴  $tan \:60\° = \frac{AB}{BC}$

substituting the values of theta & AB

$\implies \sqrt{3} = \frac{8}{BC}$

$\implies BC = \frac{8}{\sqrt{3}}$

∴ BC = AD = $\frac{8}{\sqrt{3}}$ ....... (ii) .... [since opposite sides of a rectangle are equal in length]

 

Considering Δ AED, we have

ED = perpendicular

AD = base

θ = 45°

∴  $tan \:45\° = \frac{ED}{AD}$

substituting from (ii)

$\implies 1 = \frac{ED}{\frac{8}{\sqrt{3} } }$

$\implies ED = \frac{8}{\sqrt{3}}$  ...... (iii)

 

Now,

The height of the tower is,

= EC

= ED + CD

substituting from (i) & (iii)

= $\frac{8}{\sqrt{3} } + 8$

= $\frac{8}{1.732} + 8$

= $4.61 + 8$

= $\bold{12.61\:m}$  

 

Thus, the  height of the tower from the ground is → 12.61 m.

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