Q)  From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60 and the angle of depression of its foot is 30. Determine the height of the tower.


From the top of a 7m 10th board PYQ ICSE IGCSE IB CBSE

Let’s draw a diagram with cable tower AB and Building PQ of 7m height. Angle of elevation to top of tower is given as 60° and angle of depression to tower’s foot is given as 30°. Let’s consider H be the height of cable tower and D be the distance between the tower and the building.

By interior angle property, we can say that:

∠PBQ = ∠CPB = 30°

Let’s start from Δ PQB, tan 30° = \frac{PQ}{BQ}

\frac{1}{\sqrt 3} = \frac{7}{D}

\therefore D = 7√3 m

Now in Δ APC, tan 60° = \frac{AC}{CP}

√3 = \frac{H-7}{7\sqrt3}

\therefore H – 7 = (√3)(7√3)

H – 7 = 21

H = 28 m

Therefore, the height of the cable tower is 28 m.

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