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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° (see Fig. 9.12). Find the height of the tower and the width of the canal. Q.11

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Sir give me the best solution for the question of class 10th ncert math of Applications of Trigonometry chapter of exercise 9.1 of question no.11, how i solve this problem in easy way 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° (see Fig. 9.12). Find the height of the tower and the width of the canal.

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  1. Given, AB is the height of the tower.

    DC = 20 m (given)

    As per given diagram, In right ΔABD,

    tan 30° = AB/BD

    1/√3 = AB/(20+BC)

    AB = (20+BC)/√3 … (i)

    Again,

    In right ΔABC,

    tan 60° = AB/BC

    √3 = AB/BC

    AB = √3 BC … (ii)

    From equation (i) and (ii)

    √3 BC = (20+BC)/√3

    3 BC = 20 + BC

    2 BC = 20

    BC = 10

    Putting the value of BC in equation (ii)

    AB = 10√3

    This implies, the height of the tower is 10√3 m and the width of the canal is 10 m.

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