Question

From the top of cliff 50 m 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 the tower is______
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Answer 1

50 m

R.E.F image 

Consider ΔDCB right angled at c,

tanx∘=CDBC=CD25...(1)

Now, consider ΔDCE right angled atc,

tanx∘=CDCE=...(2) 

equating eqn (1) & (2), 

⇒CD25=CDCE

⇒CE=25

∴ Total ht. of tower is (25+25) m 

(∵ BE is ht.of tower such that BC+CE & BC = AD = 25 ⇒25+CE )

=50m Ans 

So Option b is correct 

Step-by-step explanation:

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