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  1. 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.

    Best Answer
    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.

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  2. Asked: In:

    Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

    Best Answer
    jivan kumar

    Let BD is 80m wide road and AB and CD are the two equal poles. Let AB = CD = x In ΔABO, AB/BO = tan 60° ⇒ x/BO = √3 ⇒ BO = x/√3 In ΔCDO, CD/DO = tan 30° ⇒ x/(80-BO) = 1/√3 ⇒ x√3 = 80 - BO ⇒ x√3 = 80 - x/√3 [Putting the value of BO] ⇒ x√3 + x/√3 = 80 ⇒ 3x + x = 80√3 ⇒ 4x = 80√3 ⇒ x = 20√3 Therefore,Read more

    Let BD is 80m wide road and AB and CD are the two equal poles.
    Let AB = CD = x
    In ΔABO,
    AB/BO = tan 60° ⇒ x/BO = √3 ⇒ BO = x/√3
    In ΔCDO,
    CD/DO = tan 30° ⇒ x/(80-BO) = 1/√3 ⇒ x√3 = 80 – BO
    ⇒ x√3 = 80 – x/√3 [Putting the value of BO]
    ⇒ x√3 + x/√3 = 80 ⇒ 3x + x = 80√3 ⇒ 4x = 80√3 ⇒ x = 20√3
    Therefore, BO = x/√3 = 20√3/√3 = 20
    DO = BD – BO = (80 – 20) m = 60 m
    Hence, the height of pole is 20√3 m and distance of poles from the point is 20m and 60m.

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  3. Asked: In:

    A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° Find the height of the tower and the width of the canal.

    Best Answer
    jivan kumar

    Let AB is TV tower and BC is width of the canal. In ΔABC, AB/BC = tan 60° ⇒ AB/BC = √3 ⇒ BC = AB√3 In ΔABD, AB/BD = tan 30° ⇒ AB/(BC+CD) = 1/√3 ⇒ AB/(AB/√3+CD) = 1/√3 [Putting the value of BC] ⇒ AB√3/(AB + 20√3) = 1/√3 ⇒ 3AB = AB + 20√3 ⇒ 2AB = 20√3 ⇒AB = 10√3 Therefore, BC = AB/√3 = 10√3/√3= 10 HenRead more

    Let AB is TV tower and BC is width of the canal. In ΔABC,
    AB/BC = tan 60° ⇒ AB/BC = √3
    ⇒ BC = AB√3
    In ΔABD,
    AB/BD = tan 30° ⇒ AB/(BC+CD) = 1/√3
    ⇒ AB/(AB/√3+CD) = 1/√3 [Putting the value of BC]
    ⇒ AB√3/(AB + 20√3) = 1/√3 ⇒ 3AB = AB + 20√3 ⇒ 2AB = 20√3 ⇒AB = 10√3
    Therefore, BC = AB/√3 = 10√3/√3= 10
    Hence, the height of the tower is 10√3m and the width of tower is 10m.

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