Question

An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30°. If the observe moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15°, find the height of the tower.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The height of tower is 27.32 m.

Step-by-step explanation:

We have drawn the diagram for your reference.

Given:

An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30°.

according to diagram,

m ∠D = 30°

Also Given:

If the observe moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15°.

So According to diagram;

m ∠C = 45°

DC = 20 m

We need to find the height of the tower.

Let height of the tower be 'h'.

Also Let distance of CB be 'x'.

Now we know that:

$tan\ 30 =\frac{AB}{DC+CB} = \frac{h}{x+20}$

but $tan\ 30=\frac{1}{\sqrt{3}}$

so;

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

Also;

$tan\ 45 = \frac{AB}{CB} = \frac{h}{x}$

But $tan\ 45 =1$

hence ;

$1=\frac{h}{x}\\\\h=x$

Now Substituting the value of 'x' in equation 1 we get;

$h\sqrt{3} = x+20\\\\h\sqrt{3} = h+20\\\\h\sqrt{3} -h= 20\\\\h(\sqrt{3}-1)=20\\ \\h=\frac{20}{(\sqrt{3}-1)} \approx 27.32\m$

Hence the height of tower is 27.32 m.