Question

the angle of elevatiom of the top tower from 200 meter away from the tower is 45 find the height of the tower​

Answer 1

Given

➝Angle of Elevation of the top tower is 45⁰

➝AC = 200m

According to question we form a triangle ΔABC

➝Where ∠A = 45⁰

We Know That

➝TanA = Perpendicular(p)/Base(b)

we get

➝TanA = BC/AC

➝AC = 200m and BC = h

Put the value

➝Tan45⁰ = h/200m

Where

➝Tan45⁰ = 1

➝1 = h/200m

➝h = 200m

Answer

➝Height of the tower is 200m

Answer 2

Given :-

Angle of elevation = 45

Length = 200 m

To Find :-

Height

Solution :-

Let us assume that

$\sf \triangle = XYZ$

So,

XZ is the length

YZ is the height

Now,

According to the question

$\sf tan 45^{\circ} = \dfrac{YZ}{XZ}$

Where

YZ = height

XZ = length

tan 45 = 1

$\sf 1 = \dfrac{YZ}{200}$

$\sf 1 \times 200 = YZ$

$\sf 200 = YZ$

Hence,

The height of tower is 200 m