Math, asked by mohamadrafi50, 7 months ago

Q.1: The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Solution:

Answers

Answered by MяƖиνιѕιвʟє
3

Given :-

  • The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°

To find :-

  • Height of the tower

Solution :-

  • Height of tower = AB

  • ∠ADB = 30°

  • ∠ACB = 60°

  • Length of DC = 40m

  • DB = DC + CB = 40 + CB

Now, In ACB

→ tan 60° = AB/CB

→ √3 = AB/CB

→ AB = √3CB -----(i)

In ADB

→ tan 30° = AB/DB

Put the value of AB

→ 1/√3 = √3CB/40 + CB

→ 40 + CB = 3CB

→ 3CB - CB = 40

→ 2CB = 40

→ CB = 40/2

→ CB = 20m

Now, height of the tower

From (i)

→ AB = √3CB

→ AB = 20√3

Put the value of 3 is 1.73

→ AB = 20 × 1.73

→ AB = 34.6 m

Hence,

  • The height of the tower is 34.6m

Extra Information :-

  • tan θ = perpendicular/base

  • cot θ = base/perpendicular

  • sin θ = perpendicular/hypotenuse

  • cosec θ = hypotenuse/perpendicular

  • cos θ = base/hypotenuse

  • sec θ = hypotenuse/base

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Answered by Anonymous
2

Solution :

Let the AB be the height of a tower.

  • DC = 40 m

DB is 40 m longer than BC.

Therefore,

  • DB = DC + BC

________________

In ∆ABC,

tan 60° = AB/BC

√3 = AB/BC (∵ tan 60° = √3)

√3 BC = AB (Taking BC to the left hand side)

BC = AB/√3..........[Equation (1)]

________________

In ∆ABD,

tan 30° = AB/BD

1/√3 = AB/BD

BD = √3 AB

DC + BC = √3 AB (∵√3 AB DB = DC + BC)

BC = √3 AB - DC

BC = √3 AB - 40 m......[Equation (2)] (∵ It is given that DC = 40 m)

__________________....

From Equation (1) and (2) we get,

BC = BC

AB/√3 = √3 AB - 40

AB = √3 (√3 AB) - 40√3 (Taking √3 to the RHS)

AB = 3 AB - 40√3

40√3 = 3AB - AB

40√3 = 2AB

AB = 40√3/2

AB = 20√3 m

or

AB = 20 (1.732) (Taking the value of √3 = 1.732)

AB = 34.64 m

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