Question

From the top of a cliff 25m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower . The height of tower is ??

Answer 1

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Answer 2

Answer:

50 m

Step-by-step explanation:

Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C

Let angle be Q and CD = 25 m

LetAB=h CE||DB

∴EC=DB=x

(suppose)EB=CD=25

∴AE=h−25

Now in right ΔCDB,

tanθ=CDDB=25x......(i)

and in right ΔCAEtanθ=AECE=h−25x......(ii)

From(i) and (ii)2

5x=h−25x

⇒25=h−25

⇒h=25+25=50

∴Height of tower=50m

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