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    the angle of elevation of top of a building from the foot of a tower is 30 degree

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    The angle of elevation of the top of the building from the foot of the tower is 30 and the angle of the top of the tower from the foot of the building is 60 . If the tower is 50 m high, find the height of the building.

    Click here👆to get an answer to your question ✍️ The angle of elevation of the top of the building from the foot of the tower is 30 and the angle of the top of the tower from the foot of the building is 60 . If the tower is 50 m high, find the height of the building.

    Question

    The angle of elevation of the top of the building from the foot of the tower is 30 and the angle of the top of the tower from the foot of the building is 60. If the tower is 50 m high, find the height of the building.

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    Updated on : 2022-09-05

    Solution Verified by Toppr

    Given height of tower CD=50m

    Let the height of the building, AB=h

    In right angled △BDC,

    ⇒  tan60 o = BD CD ​ ⇒ 3 ​ = BD 50 ​ ⇒  BD= 3 ​ 50 ​ m

    In right angled △ABD,

    ⇒  tan30 o = BD AB ​ ⇒ 3 ​ 1 ​ = 3 ​ 50 ​ h ​ ∴ 3 ​ 1 ​ = 50 3 ​ h ​ ∴  h= 3 50 ​ =16.66m

    ∴  The height of the building is 16.66m.

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    The angle of elevation of the top of the building from the foot of the tower is 30 and the angle of ...

    Free solutions for R D Sharma Solutions - Mathematics - Class 10 Chapter 12 - Some Applications Of Trigonometry Some Applications Of Trigonometry Exercise 12.1 question 33. These explanations are written by Lido teacher so that you easily understand even the most difficult concepts

    RD Sharma Solutions Class 10 Mathematics Solutions for Some Applications Of Trigonometry Exercise 12.1 in Chapter 12 - Some Applications Of Trigonometry

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    The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

    Answer:

    Solution:

    A common notion in trigonometry, specifically, is the angle of elevation, which has to do with height and distance. It is described as an angle formed by the horizontal plane and an oblique line between the observer's eye and a target above it.

    Let AB be the building and CD be the tower.

    Given,

    The angle of elevation of the top of the building from the foot of the tower is 30o.

    And, the angle of elevation of the top of the tower from the foot of the building is 60o.

    Height of the tower = CD = 50 m

    From the fig. we have

    In ΔCDB, CD/ BD = tan 60o 50/ BD = √3 BD = 50/√3 …. (i) Next in ΔABD, AB/ BD = tan 30o AB/ BD = 1/√3 AB = BD/ √3

    AB = 50/√3/ (√3) {From (i)}

    AB = 50/3

    Therefore, the height of the building is 50/3 m.A ladder 15 metres long reaches the top of a vertical wall. If the ladder makes an angle of 60

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    Ex 9.1, 9

    Ex 9.1 , 9 The angle of elevation of the top of a building from the foot of the tower is 30 and the angle of elevation of the top of the tower from the foot of the building is 60 . If the tower is 50 m high, find the height of the building. Let building be AB & tower be CD Given Height of

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    EX 9.1, 9

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    Ex 9.1 , 9 The angle of elevation of the top of a building from the foot of the tower is 30 and the angle of elevation of the top of the tower from the foot of the building is 60 . If the tower is 50 m high, find the height of the building. Let building be AB & tower be CD Given Height of the tower = 50 m Hence, CD = 50m Angle of elevation of top of building from foot of tower = 30 Hence, ACB = 30 Angle of elevation of top of tower from foot of building = 60 Hence, DBC = 60 We need to find height of building i.e. AB Since building & tower are perpendicular to ground ABC = 90 & DCB = 90 Similarly, In a right angle triangle ABC, tan C = ( " " )/( " " ) tan 30 = (" " )/ (" " 1)/ 3 = (" " )/((" " 50)/ 3) (" " 1)/ 3 = AB 3/50 (" " 1)/ 3 50/ 3 = AB AB = (" " 1)/ 3 50/ 3 AB = (" " 50)/3 m Hence, height of building = (" " 50)/3 m

    Next: Ex 9.1, 10 →

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