Trigonometry


  1. The angles of elevation of an aeroplane flying vertically above the ground, as observed from the two consecutive stones, 1 km apart; are 45° and 60° aeroplane from the ground is :
    1. (√3 + 1) km.
    2. (√3 + 3) km.
    3. 1
      (√3 + 1) km.
      2
    4. 1
      (√3 + 3) km.
      2

  1. View Hint View Answer Discuss in Forum


    Two consecutive kilometre stones ⇒ C and D
    ∠ADB = 45°; ∠ACB = 60°
    CD = 1 km.
    AB = height of plane = h metre
    BC = x metre (let)
    In ∆ABC,

    tan60° =
    AB
    BC

    ⇒ √3 =
    h
    x

    ⇒ h = √3x metre ..... (i)
    In ∆ABD
    tan45° =
    AB
    BD

    ⇒ 1 =
    h
    x + 1

    ⇒ h = x + 1
    ⇒ h =
    h
    + 1
    3

    [From equation (i)]
    ⇒ h -
    h
    = 1
    3

    3h - h
    = 1
    3

    ⇒ (√3 - 1)h = √3
    ⇒ h =
    3
    3 - 1

    ⇒ h =
    3(√3 + 1)
    (√3 - 1)(√3 + 1)

    ⇒ h =
    3(√3 + 1)
    2

    h =
    (3 + √3)
    metre
    2

    Correct Option: D


    Two consecutive kilometre stones ⇒ C and D
    ∠ADB = 45°; ∠ACB = 60°
    CD = 1 km.
    AB = height of plane = h metre
    BC = x metre (let)
    In ∆ABC,

    tan60° =
    AB
    BC

    ⇒ √3 =
    h
    x

    ⇒ h = √3x metre ..... (i)
    In ∆ABD
    tan45° =
    AB
    BD

    ⇒ 1 =
    h
    x + 1

    ⇒ h = x + 1
    ⇒ h =
    h
    + 1
    3

    [From equation (i)]
    ⇒ h -
    h
    = 1
    3

    3h - h
    = 1
    3

    ⇒ (√3 - 1)h = √3
    ⇒ h =
    3
    3 - 1

    ⇒ h =
    3(√3 + 1)
    (√3 - 1)(√3 + 1)

    ⇒ h =
    3(√3 + 1)
    2

    h =
    (3 + √3)
    metre
    2


  1. Two men standing on same side of a pillar 75 metre high, observe the angles of elevation of the top of the pillar to be 30° and 60° respectively. The distance between two men is :
    1. 100 √3 metre
    2. 100 metre
    3. 50 √3 metre
    4. 25 √3 metre

  1. View Hint View Answer Discuss in Forum


    AB = Height of pole = 75 metre
    C and D ⇒ positions of persons
    Let, BC = x metre, BD = y metre
    ∆ACB = 60°; ∆ADB = 30°
    In ∆ABC,

    tan60° =
    AB
    BC

    ⇒ √3 =
    75
    x

    ⇒ x =
    75
    = 25 √3 metre
    3

    In ∆ABD
    tan30° =
    AB
    BD

    1
    =
    75
    3y

    ⇒ y = 75 √3 metre
    ∴ CD = y – x = (75 √3 - 25 √3) metre CD = 50√3 metre

    Correct Option: C


    AB = Height of pole = 75 metre
    C and D ⇒ positions of persons
    Let, BC = x metre, BD = y metre
    ∆ACB = 60°; ∆ADB = 30°
    In ∆ABC,

    tan60° =
    AB
    BC

    ⇒ √3 =
    75
    x

    ⇒ x =
    75
    = 25 √3 metre
    3

    In ∆ABD
    tan30° =
    AB
    BD

    1
    =
    75
    3y

    ⇒ y = 75 √3 metre
    ∴ CD = y – x = (75 √3 - 25 √3) metre CD = 50√3 metre



  1. From a point P on a level ground, the angle of elevation to the top of the tower is 30°. If the tower is 100 metre high, the distance of point P from the foot of the tower is (Take √3 = 1.73)
    1. 149 metre
    2. 156 metre
    3. 173 metre
    4. 188 metre

  1. View Hint View Answer Discuss in Forum


    Let, AB = Height of tower = 100 metre
    ∆ACB = 30°
    In ∆ABC,

    tan30° =
    AB
    BC

    1
    =
    100
    3BC

    ⇒ BC = 100√3 metre = (100 × 1.73) metre = 173 metre

    Correct Option: C


    Let, AB = Height of tower = 100 metre
    ∆ACB = 30°
    In ∆ABC,

    tan30° =
    AB
    BC

    1
    =
    100
    3BC

    ⇒ BC = 100√3 metre = (100 × 1.73) metre = 173 metre


  1. If the angle of elevation of the top of a pillar from the ground level is raised from 30° to 60°, the length of the shadow of a pillar of height 50 √3 will be decreased by
    1. 60 metre
    2. 75 metre
    3. 100 metre
    4. 50 metre

  1. View Hint View Answer Discuss in Forum


    AB = Height of pole = 50 √3 metre
    BC = Length of shadow = x metre
    When, ∠ACB = 30°
    BD = Length of shadow = y metre
    when, ∠ADB = 60°
    In ∆ABC,

    tan30° =
    AB
    BC

    1
    =
    50 √3
    3x

    ⇒ x = 50√3 × √3 = 150 metre
    In ∆ABD
    tan60° =
    AB
    BD

    ⇒ √3 =
    50 √3
    y

    ⇒ y =
    50 √3
    = 50 metre
    3

    ∴ CD = x – y = 150 – 50 = 100 metre

    Correct Option: C


    AB = Height of pole = 50 √3 metre
    BC = Length of shadow = x metre
    When, ∠ACB = 30°
    BD = Length of shadow = y metre
    when, ∠ADB = 60°
    In ∆ABC,

    tan30° =
    AB
    BC

    1
    =
    50 √3
    3x

    ⇒ x = 50√3 × √3 = 150 metre
    In ∆ABD
    tan60° =
    AB
    BD

    ⇒ √3 =
    50 √3
    y

    ⇒ y =
    50 √3
    = 50 metre
    3

    ∴ CD = x – y = 150 – 50 = 100 metre



  1. Find the angular elevation of the Sun when the shadow of a 15 metre long pole is (15 / √3) metre.
    1. 45°
    2. 60°
    3. 30°
    4. 90°

  1. View Hint View Answer Discuss in Forum


    A' ⇒ Position of sun
    AB = Height of pole = 15 metre

    BC = Length of shadow =
    15
    metre
    3

    ∴ tanθ =
    AB
    =
    15
    = √3
    BC(15 / √3)

    ⇒ tanθ = tan60°
    ⇒ θ = 60°

    Correct Option: B


    A' ⇒ Position of sun
    AB = Height of pole = 15 metre

    BC = Length of shadow =
    15
    metre
    3

    ∴ tanθ =
    AB
    =
    15
    = √3
    BC(15 / √3)

    ⇒ tanθ = tan60°
    ⇒ θ = 60°