Question

A man is standing at some distance from a 30 m high tower. The angle of elevation of the top of the tower increases from 30° to 60° as he walks towards the tower. Find the distance he walked towards the tower.​

Answer 1

Step-by-step explanation:

first find AC using 'Tan A' asA=30

then find A'C same as first

then subtract A'C from AC

Answer 2

Answer:

Distance he walked towards the tower is 20√3 m.

Step-by-step explanation:

Given :

• Height of tower is 30 m.
• Angle of elevation of top of tower increasing from 30° to 60°.

To find :

• Distance he walked towards the tower.

Solution :

Let, AB be height of tower.

BD be the distance between tower and man, till where he is making angle of 30°.

CD be distance he walked towards the tower.

So,

$\longrightarrow \sf tan \: 30\degree = \dfrac{AB}{BD}$

$\longrightarrow \sf \dfrac{1}{3} = \dfrac{30}{BD}$

$\longrightarrow \pmb{\sf BD = 30 \sqrt{3}}$

And,

$\longrightarrow \sf tan \: 60\degree = \dfrac{AB}{BC}$

$\longrightarrow \sf \sqrt{3} = \dfrac{30}{BC}$

$\longrightarrow \sf BC \times \sqrt{3} = 30$

$\longrightarrow \sf BC = \dfrac{30}{\sqrt{3}}$

$\longrightarrow \sf BC = \dfrac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$\longrightarrow \sf BC = \dfrac{30 \sqrt{3}}{3}$

$\longrightarrow \pmb{\sf BC = 10 \sqrt{3}}$

Now,

→ CD = 30√3 - 10√3

→ CD = 20√3.

∴ Distance he walked towards the tower is 20√3 m.