Question

man observes top of Tower at an angle of elevation of 30° when he walked 40 metres towards the tower the angle of elevation is changed into 60 degree find the height of the tower and the distance from the first observation point to our​

Answer 1

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Answer 2

Height of tower=$20\sqrt 3m$

The distance from the first observation point to foot of tower=60 m

Step-by-step explanation:

Let height of tower=h and BC= x

AB=h

BD=BC+CD=x+40

In triangle ABC

$\theta=60^{\circ}$

$\frac{AB}{BC}=tan60^{\circ}$

Using

$\frac{Perpendicular\;side}{Base}=tan\theta$

$\frac{h}{x}=\sqrt 3$

Using

$tan60^{\circ}=\sqrt 3$

$h=x\sqrt 3$......(1)

In triangle ABD

$\frac{AB}{BD}=tan30^{\circ}$

$\frac{h}{x+40}=\frac{1}{\sqrt 3}$

Using

$tan30^{\circ}=\frac{1}{\sqrt 3}$

Substitute the value

$\frac{x\sqrt 3}{x+40}=\frac{1}{\sqrt 3}$

$3x=x+40$

$3x-x=40$

$2x=40$

$x=\frac{40}{2}=20 m$

BD=20+40=60 m

Substitute the values of x

$h=20\sqrt 3 m$

Height of tower=$20\sqrt 3m$

Hence, the distance from the first observation point to foot of tower=60 m

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