Question

From the top of a cliff 25 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 ​

Answer 1

Step-by-step explanation:

R.E.F image

Consider ΔDCB right angled at c,

tanx

∘

=

CD

BC

=

CD

25

...(1)

Now, consider ΔDCE right angled atc,

tanx

∘

=

CD

CE

=...(2)

equating eq

n

(1) & (2),

⇒

CD

25

=

CD

CE

⇒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