Question

The angle of elevation of the top of a tower from a point situated at 100 m far from the foot of tower is 30°.find height of the tower

Answer 1

Answer:

the height of the tower is 100/$\sqrt{3}$

Step-by-step explanation:

Let,

The height of tower (ab)=x

The distance from the tower (bc)=100m

and angle of elevation angle acb = 30 degrees

Now,

Tan 30 degrees=ab/bc

=>1/$\sqrt{3}$=x/100

=>100/$\sqrt{3}$=x

Hence, Height of tower ac=x=100/$\sqrt{3}$

Answer 2

Step-by-step explanation:

See figure of this question in attachment

Let AB is Tower

BC is distance between Tower & A point

Angle of elevation is 30°

Theta = 30°

Tan Theta. = tan 30°

tan 30° = BC/AB

1/root 3 = 100/AB

AB = 100root 3

root 3 = 1.73

height of the tower is 173 meter

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