Question

CAERUIUE -
.
A TV tower stands vertically on the side of a road. From a point on the other side
directly opposite to the tower, the angle of elevation of the top oftower is 60°. From
another point 10 m away from this point, on the line joining this point to the foot of the
tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower
and the width of the road.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given :

CD = 10 m ,

∠ACB = 60°

∠ADB = 30°

To find : AB and BD

solution :

in triangle ABC

$\tan 60^\circ = \frac{h}{x}\\\\\sqrt{3}=\frac{h}{x}\\\\h = x\sqrt{3}..(1)$

in triangle ABD

$\tan 30^\circ = \frac{h}{10+x}\\\\\frac{1}{\sqrt{3}}=\frac{h}{10+x}\\\\h\sqrt{3} = 10+x..(2)$

from 1 and 2

$x\sqrt{3}= \frac{10+x}{\sqrt{3} } \\\\3x= 10+x\\\\x = 5$

and

put the value of x we get

$h = 5\sqrt{3}$

thus , The height and width of the tower is 5√3 m and 5 m

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