Trigonometry


  1. The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases to 60°. The height of the tower is









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    Let PQ = h metre and BQ = x metre.
    From ∆ APQ,

    tan30° =
    h
    x + 20

    1
    =
    h
    3x + 20

    ⇒ √3h = x + 20 ...........(i)
    From ∆ PQB,
    tan 60° =
    PQ
    =
    h
    BQx

    ⇒ √3 =
    h
    ⇒ h = √3x
    x

    ⇒ x =
    1
    h .......(ii)
    3

    ∴ √3h =
    1
    h + 20
    3

    [From equation (i) and (ii)]
    ⇒ 3h – h = 20 √3
    ⇒ 2h = 20 √3
    ∴ h = 10 √3 metre

    Correct Option: C


    Let PQ = h metre and BQ = x metre.
    From ∆ APQ,

    tan30° =
    h
    x + 20

    1
    =
    h
    3x + 20

    ⇒ √3h = x + 20 ...........(i)
    From ∆ PQB,
    tan 60° =
    PQ
    =
    h
    BQx

    ⇒ √3 =
    h
    ⇒ h = √3x
    x

    ⇒ x =
    1
    h .......(ii)
    3

    ∴ √3h =
    1
    h + 20
    3

    [From equation (i) and (ii)]
    ⇒ 3h – h = 20 √3
    ⇒ 2h = 20 √3
    ∴ h = 10 √3 metre


  1. The angle of elevation of the top of a building from the top and bottom of a tree are x and y respectively. If the height of the tree is h metre, then (in metre) the height of the building is









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    CD = tree = h metre
    Let AB = building = a metre
    & BC = ED = b metre
    ∴ From ∆ AED,

    tanx =
    AE
    ED

    ⇒ tan x =
    a - h
    b

    ⇒ b = (a – h) cot x ..........(i)
    From ∆ ABC,
    tany =
    AB
    BC

    ⇒ tan y =
    a
    b

    ⇒ b = a cot y ...........(ii)
    From equations (i) and (ii),
    (a – h) cot x = a cot y
    ⇒ a cot x – h cot x = a cot y
    ⇒ h cot x = a (cot x – cot y)
    ⇒ a =
    h cot x
    cot x - cot y

    Correct Option: C


    CD = tree = h metre
    Let AB = building = a metre
    & BC = ED = b metre
    ∴ From ∆ AED,

    tanx =
    AE
    ED

    ⇒ tan x =
    a - h
    b

    ⇒ b = (a – h) cot x ..........(i)
    From ∆ ABC,
    tany =
    AB
    BC

    ⇒ tan y =
    a
    b

    ⇒ b = a cot y ...........(ii)
    From equations (i) and (ii),
    (a – h) cot x = a cot y
    ⇒ a cot x – h cot x = a cot y
    ⇒ h cot x = a (cot x – cot y)
    ⇒ a =
    h cot x
    cot x - cot y



  1. The distance between two pillars of length 16 metres and 9 metres is x metres. If two angles of elevation of their respective top from the bottom of the other are complementary to each other, then the value of x (in metres) is









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    From ∆ABC,

    tan (90° – θ) =
    16
    x

    ⇒ cotθ =
    16
    ........(i)
    x

    From ∆BCD,
    tanθ =
    9
    ........(ii)
    x

    ∴ tanθ . cotθ =
    9
    ×
    16
    xx

    ⇒ x2 = 16 × 9 [∵ tanθ cotθ =1]
    ⇒ x = 4 × 3 = 12 metre

    Correct Option: C


    From ∆ABC,

    tan (90° – θ) =
    16
    x

    ⇒ cotθ =
    16
    ........(i)
    x

    From ∆BCD,
    tanθ =
    9
    ........(ii)
    x

    ∴ tanθ . cotθ =
    9
    ×
    16
    xx

    ⇒ x2 = 16 × 9 [∵ tanθ cotθ =1]
    ⇒ x = 4 × 3 = 12 metre


  1. At a point on a horizontal line through the base of a monument, the angle of elevation of the top of the monument is found to be such that its tangent is 1/5. On walking 138 metres towards the monument the secant of the angle of elevation is found to be
    193
    .
    12
    The height of the monument (in metre) is









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    AB = monument = h metre
    DC = 138 metre
    BD = x metre

    tan α =
    1
    5

    sec β =
    193
    12

    ∴ tan β = √sec2 β - 1

    ∴ From ∆ABC,
    tan α =
    AB
    BC

    1
    =
    h
    5x + 138

    ⇒ h =
    x + 138
    5

    ⇒ 5h = x + 138 .............. (i)
    From ∆ ABD,
    tan β =
    h
    7
    =
    h
    x12x

    ⇒ x =
    12h
    .........(ii)
    7

    ∴ 5h =
    12h
    + 138 (By (i) & (ii)
    7

    ⇒ 35h – 12h = 138 × 7
    ⇒ 23 h = 138 × 7
    ⇒ h =
    138 × 7
    = 42 metre
    23

    Correct Option: C


    AB = monument = h metre
    DC = 138 metre
    BD = x metre

    tan α =
    1
    5

    sec β =
    193
    12

    ∴ tan β = √sec2 β - 1

    ∴ From ∆ABC,
    tan α =
    AB
    BC

    1
    =
    h
    5x + 138

    ⇒ h =
    x + 138
    5

    ⇒ 5h = x + 138 .............. (i)
    From ∆ ABD,
    tan β =
    h
    7
    =
    h
    x12x

    ⇒ x =
    12h
    .........(ii)
    7

    ∴ 5h =
    12h
    + 138 (By (i) & (ii)
    7

    ⇒ 35h – 12h = 138 × 7
    ⇒ 23 h = 138 × 7
    ⇒ h =
    138 × 7
    = 42 metre
    23



  1. The angle of elevation of the top of a tower from two points A and B lying on the horizontal through the foot of the tower are respectively 15° and 30°. If A and B are on the same side of the tower and AB = 48 metre, then the height of the tower is :









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    Tower = PQ = h metre
    QB = x metre
    From ∆ APQ,

    tan15° =
    h
    x + 48

    2 – √3 =
    h
    ....(i)
    x + 48

    [∵ tan 15° = tan (45° – 30°)
    = = 1 -
    1
    tan 45° - tan 30°3
    1 + tan 45°tan 30°1 +
    1
    3

    or
    3 - 1
    ×
    3 - 1
    3 + 13 - 1

    =
    4 - 2√3
    = 2 - √3 ]
    2

    From ∆PQB,
    tan30° =
    h
    x

    1
    =
    h
    3x

    ⇒ √3h = x .............(ii)
    ⇒ 2 - √3 =
    h
    3h + 48

    ⇒ 2√3h - 3h + (2 - √3) 48 = h
    ⇒ h + 3h - 2√3h
    =(2 - √3) × 48
    ⇒ 2h(2 - √3) = 48 × ( 2 - √3)
    ⇒ h =
    48
    = 24 metre
    2

    Correct Option: B


    Tower = PQ = h metre
    QB = x metre
    From ∆ APQ,

    tan15° =
    h
    x + 48

    2 – √3 =
    h
    ....(i)
    x + 48

    [∵ tan 15° = tan (45° – 30°)
    = = 1 -
    1
    tan 45° - tan 30°3
    1 + tan 45°tan 30°1 +
    1
    3

    or
    3 - 1
    ×
    3 - 1
    3 + 13 - 1

    =
    4 - 2√3
    = 2 - √3 ]
    2

    From ∆PQB,
    tan30° =
    h
    x

    1
    =
    h
    3x

    ⇒ √3h = x .............(ii)
    ⇒ 2 - √3 =
    h
    3h + 48

    ⇒ 2√3h - 3h + (2 - √3) 48 = h
    ⇒ h + 3h - 2√3h
    =(2 - √3) × 48
    ⇒ 2h(2 - √3) = 48 × ( 2 - √3)
    ⇒ h =
    48
    = 24 metre
    2