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 foot of the tower find the height of the tower.

Answer 1

let CD be the Ht of the tower,
and AB be the Cliff
then, AB = 25m. Draw BE ⊥ CD
Let ∠EBD=∠ACB=α
now, DE/BE=tanα and AB/AC=tanα
∴ DE/BE=AB/AC
so, DE=AB  [∵BE=AC]
now, CD=CE+DE
             =AB+AB   [from above step and CE=AB because they are opp                                        sides of rectangle]
             = 2AB
             =2(25)
             = 50m

Answer 2

answer is given

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