Q. 6

# In Fig. 12.58, wh

Let x be the angle of depression of object A from the point O2

And y be the angle of depression of object A from the point O1

In Δ AO1C

O1AC + AC O1 + ACO1 = 180° (angle sum property)

O1AC + 90° + 60° = 180°

O1AC = 30°

Through O1 , draw O1M AC

And through O2 draw O2N AC

Now O1M AC and AO1 is transversal

O1AC = y = 30° = MO1A (vertically opposite s)

Similarly, O2N AC and AO2 is transversal

NO2 A = x = 45° = O2 AB (vertically opposite angles)

Angles of depression are 30° and 45°

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