Question

The angle of elevation of the top of a tower at a point on the level ground is 30 degree . After walking distance of 100m towards the foot of the tower along the horizontal line through the foot of the tower on the same level ground the angle of elevation of the tower is60 degree . Find the height of the tower

Answer 1

Answer:

86.6 m  

Step-by-step explanation:

Refer the attached figure .

Height of tower AB = h  

∠ACB = 60°

∠ADB = 30°

Walking distance of 100m towards the foot of the tower along the horizontal line through the foot of the tower on the same level i.e. CD = 100m

Let BC = x  

So, BD = x+100

In ΔABC

Using trigonometric ratios

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

$tan 60 ^{\circ} = \frac{AB}{BC}$

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

$\sqrt{3}x = h$ -1

In ΔABD

Using trigonometric ratios

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

$tan 30 ^{\circ} = \frac{AB}{BD}$

$\frac{1}{\sqrt{3}}= \frac{h}{x+100}$

$\frac{x+100}{\sqrt{3}} = h$ --2

Equating 1 and 2  

$\frac{x+100}{\sqrt{3}}  = \sqrt{3}x$  

$x+100 =3x$  

$100 =2x$  

$50 =x$  

Substitute the values of x in 1  

$\sqrt{3}50 = h$

$86.6 m = h$

Hence the height of the tower is 86.6 m  

Answer 2

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