Question

the angle of elevation of the top of an unfinished tower at a point distant 120m from its base is 45 degree.how much higher must the tower to be raised so that its angle of elevation at the same point may be 60

Answer 1

Answer:

Let AB be the tower initially and D be the point on the ground.

DB=120m, angle ADB = 45 degree

CB be the final height & angle CDB = 60 degree.

A/Q In ∆CDB

tan60 = CB ÷ 120

√3 = CB / 120

Therefore, CB = 120√3m

now, In ∆ ADB

tan45 = AB / 120

1 = AB / 120

Therefore, AB = 120m

CB = AB + AC

CB - AB = AC

120√3 - 120 = AC

AC = 120 ( √3 - 1 ) m

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