Home/ jivan kumar/Best Answers
  1. Asked: In:

    The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

    jivan kumar

    Let AQ is tower and C and D are the two point 4 m and 9 m away from the foot of tower. The angle of elevations are complementary. So if one angle is θ, then the other will be 90- θ. In ΔAQR, AQ/QR = tan θ ⇒ AQ/4 = tan θ ...(1) In ΔAQS, AQ/QS = tan (90 - θ) ⇒ AQ/9 = cot θ ...(2) Multiplying the equatRead more

    Let AQ is tower and C and D are the two point 4 m and 9 m away from the foot of tower.
    The angle of elevations are complementary. So if one angle is θ, then the other will be 90- θ.
    In ΔAQR,
    AQ/QR = tan θ
    ⇒ AQ/4 = tan θ …(1)
    In ΔAQS,
    AQ/QS = tan (90 – θ)
    ⇒ AQ/9 = cot θ …(2)
    Multiplying the equations (1) and (2), we have
    (AQ/4)(AQ/9) = tan θ. cot θ
    ⇒ AQ²/36 = 1
    ⇒ AQ² = 36
    ⇒ AQ = ±6
    Height can’t be negative. Hence, the height of the tower is 6 m.

    See less
    • 1
  2. Asked: In:

    From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

    jivan kumar

    Let AB is building and CD is cable tower. In ΔABD, AB/BD = tan 45° ⇒ 7/BD = 1 ⇒ BD = 7 In ΔACE, AC = BD = 7 m CE/AE = tan 60° ⇒ CE/7 = √3 ⇒ CE = 7√3 Therefore, CD = CE + ED = 7√3 + 7 = 7(√3 + 1) Hence, the height of the cable tower is 7(√3 + 1) m.

    Let AB is building and CD is cable tower.
    In ΔABD,
    AB/BD = tan 45° ⇒ 7/BD = 1 ⇒ BD = 7
    In ΔACE, AC = BD = 7 m
    CE/AE = tan 60° ⇒ CE/7 = √3 ⇒ CE = 7√3
    Therefore, CD = CE + ED = 7√3 + 7 = 7(√3 + 1)
    Hence, the height of the cable tower is 7(√3 + 1) m.

    See less
    • 1
  3. Asked: In:

    From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

    jivan kumar

    Let BC is Building and AB is transmission tower. In ΔBCD, BC/CD = tan 45° ⇒ 20/CD = CD = 1 In ΔACD, AC/ CD = tan 60° ⇒ AB+BC/CD = √3 ⇒ (AB+20)/20 = √3 ⇒ AB + 20 = 20√3 ⇒ AB = 20(√3- 1) Hence, the height of the transmission towwer is 20(√3- 1)m.

    Let BC is Building and AB is transmission tower.
    In ΔBCD,
    BC/CD = tan 45° ⇒ 20/CD = CD = 1
    In ΔACD,
    AC/ CD = tan 60° ⇒ AB+BC/CD = √3 ⇒ (AB+20)/20 = √3
    ⇒ AB + 20 = 20√3 ⇒ AB = 20(√3- 1)
    Hence, the height of the transmission towwer is 20(√3- 1)m.

    See less
    • 2
  4. Asked: In:

    A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

    jivan kumar

    Let BC is Pedestal and AB is 1.6 m high statue. In ΔBCD, BC/CD = tan 45° ⇒ BC/CD = 1 ⇒ BC = CD In ΔACD, AB+BC/CD = tan 60° ⇒ AB+BC/BC = √3 ⇒ 1.6 + BC/BC = √3 ⇒ BC + 1.6 = BC√3 ⇒ 1.6 = BC (√3 - 1) ⇒ BC = 1.6/(√3-1) × (√3+1)/(√3+1) = 1.6(√3+1)/2 = 0.8(√3+1) Hence, the height of the pedestal is 0.8(√3+Read more

    Let BC is Pedestal and AB is 1.6 m high statue.
    In ΔBCD,
    BC/CD = tan 45° ⇒ BC/CD = 1 ⇒ BC = CD
    In ΔACD,
    AB+BC/CD = tan 60° ⇒ AB+BC/BC = √3
    ⇒ 1.6 + BC/BC = √3
    ⇒ BC + 1.6 = BC√3
    ⇒ 1.6 = BC (√3 – 1)
    ⇒ BC = 1.6/(√3-1) × (√3+1)/(√3+1) = 1.6(√3+1)/2
    = 0.8(√3+1)
    Hence, the height of the pedestal is 0.8(√3+1)m.

    See less
    • 1
  6. Asked: In:

    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.

    jivan kumar

    Let CD is 50m high tower and AB is building. In ΔCDB, CD/BD = tan 60° ⇒ 50/BD = √3 ⇒ BD = 50/√3 In ΔABD, AB/BD = tan 30° ⇒ AB/(50√3) = 1/√3 ⇒ AB = 50/√3 × 1/√3 = 50/ 3 = 16 (2/3) Hence, the height of the building is 16(2/3)m.

    Let CD is 50m high tower and AB is building.
    In ΔCDB,
    CD/BD = tan 60° ⇒ 50/BD = √3 ⇒ BD = 50/√3
    In ΔABD,
    AB/BD = tan 30° ⇒ AB/(50√3) = 1/√3 ⇒ AB = 50/√3 × 1/√3 = 50/ 3 = 16 (2/3)
    Hence, the height of the building is 16(2/3)m.

    See less
    • 1