Physics, asked by dushal1678, 5 months ago

From the top of a 7m high building the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45° .​

Answers

Answered by ExᴏᴛɪᴄExᴘʟᴏʀᴇƦ
55

Answer

  • Height of the tower is 7(√3+1) m

Explanation

Given

  • From the top of a 7 m High building the angle of elevation to the top is 60°
  • Angel of depression to the foot is 45°

To Find

  • Height of the tower

Solution

  • Refer the diagram for the Diagram? Here we shall first use the tan function to find the horizontal length first, then use the same on ∆AED to find the height of DE. Then the total height will be DE + EC

✭ In ∆ABC

→ tan θ = Opposite/Adjacent

→ tan 45° = AB/BC

[tan 45°] = 1

→ 1 = 7/y

→ 1 × y = 7

→ y = 7 m

✭ In ∆AED

→ tan θ = Opposite/Adjacent

→ tan 60° = DE/AE

[tan 60° = √3]

→ √3 = x/y

→ √3 = x/7

→ √3 × 7 = x

→ x = 7√3

✭ Height of the tower

→ Height (Tower) = DE+EC

→ Height (Tower) = 7√3 + 7

→ Height (Tower) = 7(√3+1) m

Attachments:
Answered by Anonymous
82

Answer:

 \underline{ \sf{ \underline{☃Given:}}}

  • From the top of a 7m high building
  • Angle of elevation of the top of a cable tower is 60°

 \underline{ \sf{ \underline{☃Find:}}}

  • Height of tower

 \underline{ \sf{ \underline{ ☃Solution:}}}

From the attachment,

Let AB be the building of height 7 m and EC be the height of tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot

{ \mapsto{ \rm{{EC = DE + CD}}}}

{ \mapsto{  \rm{BC = AD}}}

{ \mapsto{ \rm{CD = AB = 7m}}}

From ΔABC,

 \:  \:  \: { \rm{tan45 =  \frac{ AB }{BC} }}

We know that ,Tan45° = 1

 \to{ \rm{1 =  \frac{7}{BC} }}

{ \to{  \rm{BC =AD =  7m }}}

 \:  \:  \: { \rm{tan60 =  \frac{DE }{AD} }}

We know that, Tan60° =√3

{ \to{ \rm{ \sqrt{3}  =  \frac{DE}{7} }}}

{ \to{ \rm{DE = 7 \sqrt{3} }}}

Height of tower :-

EC = DE + CD

= 7√3 + 7

= 7(√3 + 1 )

Explanation:

Therefore,Height of tower = 7 (3 + 1) m

Attachments:
Similar questions