Height and Distance

Height and Distance

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  1. The angle of elevation of the top of a hill from each of the vertices A, B, C of a horizontal triangle is α. The height of the hill is :








  1. View Hint View Answer Discuss in Forum

    As per the given details of above question , we have
    The distance of the foot from each vertex = h cotα.
    ∴ The foot is at the circumcentre of the triangle.
    ∴ R = h cotα

    Correct Option: B

    As per the given details of above question , we have
    The distance of the foot from each vertex = h cotα.
    ∴ The foot is at the circumcentre of the triangle.
    ∴ R = h cotα

    ∴ R = h tanα = atanα = a tanα . cosecα .
    2sinα2

  1. The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45°. If the elevation of the top at the same point is to be 60°, the tower must be raised to a height :








  1. View Hint View Answer Discuss in Forum

    Let us draw the figure from the given question.
    Given :- OB = h + x

    h + x = tan 60° = √3
    120

    h + x= √3 (120) .


    Correct Option: B

    Let us draw the figure from the given question.
    Given :- OB = h + x
    In figure , we have

    h + x = tan 60° = √3
    120

    h + x= √3 (120) ...........( 1 )

    Also, h = tan 45° = 1 .
    120

    ∴ h = 120 m

    From eq. ( 1 ) ,
    ∴ h + x = 120 + x = 120 √3
    ∴ x = 120 ( √3 - 1 ) m .


  1. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retires 40 m from the bank, he finds the angle to be 30°. The breadth of the river is :








  1. View Hint View Answer Discuss in Forum

    Let us draw the figure from the given question.
    Let, OA denote the breadth of the river.
    Given :- AB = 40 m


    Correct Option: C

    Let us draw the figure from the given question.
    Let, OA denote the breadth of the river.
    Given :- AB = 40 m
    In triangle OAP , we have

    OP= tan60° = √3
    OA

    ∴OP = √3 OA.

    Also, In triangle OBP , we have OP = tan 30° = 1
    OA + 40√3

    ∴ OA + 40 = √3 OP = √3 ( √3 OA ) = 3 OA.

    ⇒ 2OA = 40 ⇒ OA = 20 m.

    Hence , the breadth of the river = 20 m .


  1. The angle of elevation of the top of a tower at a point G on the ground is 30°. On walking 20 m towards the tower, the angle of elevation becomes 60°. The height of the tower is equal to :








  1. View Hint View Answer Discuss in Forum

    Let us draw the figure from the given question.
    Let, AB = h be the height of the tower. Let, GA = x.


    Correct Option: D

    Let us draw the figure from the given question.
    Let, AB = h be the height of the tower. Let, GA = x and GH = 20 m.
    In triangle BGA ,

    Then, h = tan30° = 1 .
    x√3

    ∴ h = x ......... ( 1 )
    √3

    In triangle BHA , Also,
    h
    		 = tan60° = √3 .
    x - 20

    ∴ h = √3 ( x - 20 )

    ∴ x = √3 ( x - 20 ) [ using ( 1 ) ]
    √3

    ⇒ x = 3 ( x - 20 ) = 3x - 60

    ⇒ 2x = 60 ⇒ x = 30 .
    Putting the value of x in eq. ( 1 ) , we get
    ∴ h = 30 = 10 √3 m .
    √3


  1. The angle of elevation of the top of a tower from a point 20 m away from its base is 45°. The height of the tower is :








  1. View Hint View Answer Discuss in Forum

    Let us draw the figure from the given question.
    Let us assume the angle of elevation ∠MŌP = 45°.
    Given :- MO = 20 m
    In triangle MOP ,

    tan45° = PM
    MO

    Clearly,h = tan45 = 1
    20


    Correct Option: B

    From the given figure , we can see that
    Let us assume the angle of elevation ∠MŌP = 45°. and The height of the tower = h
    Given :- MO = 20 m
    In triangle MOP ,

    tan45° = PM
    MO

    ⇒ h = tan45° = 1
    20

    ∴ h = 20 m