Trigonometry


  1. If the elevation of the Sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 metre high, is









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    AB = Height of pole = 15 metre
    ∠ACB = 60°; ∠ADB = 30°
    In ∆ABC,

    tan 60° =
    AB
    ⇒ √3 =
    15
    BCBC

    ⇒ BC =
    15
    = 5√3 metre
    3

    In ∆ABD,
    tan 30° =
    AB
    BD

    1
    =
    15
    3BD

    ⇒ BD = 15 √3 metre
    ∴ Required difference
    = BD – BC = (15√3 - 5√3) metre
    = 10 √3 metre

    Correct Option: C


    AB = Height of pole = 15 metre
    ∠ACB = 60°; ∠ADB = 30°
    In ∆ABC,

    tan 60° =
    AB
    ⇒ √3 =
    15
    BCBC

    ⇒ BC =
    15
    = 5√3 metre
    3

    In ∆ABD,
    tan 30° =
    AB
    BD

    1
    =
    15
    3BD

    ⇒ BD = 15 √3 metre
    ∴ Required difference
    = BD – BC = (15√3 - 5√3) metre
    = 10 √3 metre


  1. On a ground, there is a vertical tower with a flagpole on its top. At a point 9 metre away from the foot of the tower, the angles of elevation of the top and bottom of the flagpole are 60° and 30° respectively. The height of the flagpole is :









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    AB = Height of tower = h metre
    AD = Height of flagstaff = x metre
    ∠BCD = 60°; ∠BCA = 30°
    BC = 9 metre
    In ∆ABC,

    tan 30° =
    AB
    BC

    1
    =
    h
    39

    ⇒ h =
    9
    = 3√3 metre
    3

    In ∆BCD,
    tan 60° =
    BD
    BC

    ⇒ √3 =
    h + x
    9

    ⇒ h + x = 9 √3
    ⇒ 3 √3 + x = 9 √3
    ⇒ x = 9 √3 – 3 √3
    = 6 √3 metre

    Correct Option: B


    AB = Height of tower = h metre
    AD = Height of flagstaff = x metre
    ∠BCD = 60°; ∠BCA = 30°
    BC = 9 metre
    In ∆ABC,

    tan 30° =
    AB
    BC

    1
    =
    h
    39

    ⇒ h =
    9
    = 3√3 metre
    3

    In ∆BCD,
    tan 60° =
    BD
    BC

    ⇒ √3 =
    h + x
    9

    ⇒ h + x = 9 √3
    ⇒ 3 √3 + x = 9 √3
    ⇒ x = 9 √3 – 3 √3
    = 6 √3 metre



  1. A telegraph post is bent at a point above the ground. Its top just touches the ground at a distance of 8 √3 metre from its foot and makes an angle of 30° with the horizontal. The height (in metre) of the post is :









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    Let telegraph pole bend at point A.
    BC = 8 √3 metre
    In ∆ABC,

    tan 30° =
    AB
    BC

    1
    =
    AB
    38√3

    ⇒ AB =
    8√3
    = 8 metre
    3

    Again, sin 30° =
    AB
    AC

    1
    =
    8
    AC2

    ⇒ AC = 2 × 8 = 16 metre
    ∴ Height of telegraph–pole
    = AB + AC
    = 8 + 16 = 24 metre

    Correct Option: D


    Let telegraph pole bend at point A.
    BC = 8 √3 metre
    In ∆ABC,

    tan 30° =
    AB
    BC

    1
    =
    AB
    38√3

    ⇒ AB =
    8√3
    = 8 metre
    3

    Again, sin 30° =
    AB
    AC

    1
    =
    8
    AC2

    ⇒ AC = 2 × 8 = 16 metre
    ∴ Height of telegraph–pole
    = AB + AC
    = 8 + 16 = 24 metre


  1. The upper part of a tree broken at a certain height makes an angle of 60° with the ground at a distance of 10 metre from its foot. The original height of the tree was









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    AB = Height of tree
    Let the tree break at point C.
    BC = x metre
    ∴ AC = CD
    ∠CDB = 60°; BD = 10 metre
    In ∆BCD,

    tan 60° =
    BC
    ⇒ √3 =
    x
    BD10

    ⇒ x = 10 √3 metre
    Again, sin 60° =
    BC
    CD

    3
    =
    10√3
    2CD

    ⇒ CD =
    2 × 10√3
    = 20 metre
    3

    ∴ Height of tree = AB
    = (20 + 10√3) metre
    = 10 (2 + √3) metre

    Correct Option: C


    AB = Height of tree
    Let the tree break at point C.
    BC = x metre
    ∴ AC = CD
    ∠CDB = 60°; BD = 10 metre
    In ∆BCD,

    tan 60° =
    BC
    ⇒ √3 =
    x
    BD10

    ⇒ x = 10 √3 metre
    Again, sin 60° =
    BC
    CD

    3
    =
    10√3
    2CD

    ⇒ CD =
    2 × 10√3
    = 20 metre
    3

    ∴ Height of tree = AB
    = (20 + 10√3) metre
    = 10 (2 + √3) metre



  1. The angle of elevation of the top of an unfinished pillar at a point 150 metres from its base is 30°. The height (in metres) that the pillar must be raised so that its angle of elevation at the same point may be 45°, is (Take, √3 = 1.732)









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    AB = incomplete pole
    BC = 150 metre
    ∠ACB = 30°
    In ∆ABC,

    tan30°
    AB
    BC

    1
    =
    AB
    3150

    ⇒ AB =
    150
    = 50√3 metre
    3

    In ∆BCD,
    tan45° =
    BD
    BC

    ⇒ 1 =
    BD
    150

    ⇒ BD = 150 metre
    ∴ AD = BD – AB
    = (150 – 50 √3 ) metre
    = 50 (3 - √3) metre
    = 50 (3 – 1.732) metre
    = (50 × 1.268) metre
    = 63.4 metre

    Correct Option: A


    AB = incomplete pole
    BC = 150 metre
    ∠ACB = 30°
    In ∆ABC,

    tan30°
    AB
    BC

    1
    =
    AB
    3150

    ⇒ AB =
    150
    = 50√3 metre
    3

    In ∆BCD,
    tan45° =
    BD
    BC

    ⇒ 1 =
    BD
    150

    ⇒ BD = 150 metre
    ∴ AD = BD – AB
    = (150 – 50 √3 ) metre
    = 50 (3 - √3) metre
    = 50 (3 – 1.732) metre
    = (50 × 1.268) metre
    = 63.4 metre