Question

The angle of elevation of the top of a tower from an observer on the ground is 60 .if the tower is 40m high how far is the observer from the foot of the Tower

Answer 1

Given :

• The angle of elevation of the top of a tower from an observer on the ground is 60°.
• Height of the tower is 40 m.

To find :

• Distance between the observer and the foot of the tower.

Solution :

According to the attachment,

• AB = Height of the tower.
• BC = Distance between the observer and the foot of the tower.

AB = 40 m.

$\sf{\angle\:DAC=60^\degree}$

• $\sf{\angle\:DAC=\angle\:ACB\:[Alternative\: angles]}$

So,

$\sf{\angle\:ACB=60^\degree}$

∆ ABC is a right triangle.

$\implies\sf{\frac{AB}{BC}=tan60^\degree}$

$\implies\sf{\frac{40}{BC}=\sqrt{3}}$

$\implies\sf{BC\sqrt{3}=40}$

$\implies\sf{BC=\frac{40}{\sqrt{3}}}$

$\implies\sf{BC=\frac{40}{1.732}}$

$\implies\sf{BC=23.09\:(approx)}$

Therefore, the distance between the observer and the foot of the tower is 23.09 m ( approx ).

Answer 2

❥ Qᴜᴇsᴛɪᴏɴ :-

The angle of elevation of the top of a tower from an observer on the ground is 60. If the tower is 40m high how far is the observer from the foot of the tower.

❥ Gɪᴠᴇɴ :-

• Height of the tower = 40m
• The angle of elevation of the top of a tower from an observer on the ground = 60°

❥ Tᴏ ғɪɴᴅ :-

• The Distance between foot of the tower & observer.

❥ Sᴏʟᴜᴛɪᴏɴ :-

Rᴇғᴇʀ ᴛʜᴇ ɪᴍᴀɢᴇ ғɪʀsᴛ

• Let AB be the height of the tower
• BC be the distance between foot of the tower & observer.
• AB = 60m
• ∠DAC = 60°
• ∠ACB = 60° also

❥ Sᴏʟᴠɪɴɢ Mᴀᴛʜᴇᴍᴀᴛɪᴄᴀʟʟʏ :-

⟾ AB/BC = tan 60°

⟾ 40/BC = √3

⟾ BC = 40/√3

⟾ BC = 40/1.732

⟾ BC = 23.09

⛬ The Distance between observer & foot of the tower = 23.09 meters