Question

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 of tower 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

ＱＵＥＳＴＩＯＮ :-

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 of tower 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.

ＳＯＬＵＴＩＯＮ :-

• Let the height of the tower = h m
• width of the road = x m
• Distance between 2 points of observation = 10 cm
• Angles of elevation from the 2 points = 60° & 30°.

Refer the image

From the figure

tan60° = h/x

=> √3 = h/x

=> h = √3x .............( 1 )

Also tan30° = h/10+x

=> 1/√3 = h/10 + x

=> h = 10+x/√3 ...........( 2 )

From equation 1 and 1

h = √3x = 10+x/√3

.°. √3 x = 10+x/√3

=> √3.√3x = 10 + x

=> 3x - x = 10

=> 2x = 10

=> x = 10/2

=> x = 5

Width of the road = 5 m

Height of the tower = √3x = 5√3 m

Answer 2

Answer:

$\huge\boxed{\fcolorbox{red}{pink}{Your's Answer}}$

In this figure; CD=10m,∠ACB=60°,∠ADB=30°,AB=? and BD=?

In △ABC,

$tan60 = \frac{h}{x}$

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

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

In △ABD,

$tan30 = \frac{h}{10 + x}$

$\frac{1}{ \sqrt{3} } = \frac{h}{10 + x}$

$h \sqrt{3} = 10 + x...........(2)$

From equation (1) and (2);

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

3x=10+x

x=5, which is the width of the road.

From (1),

$h = 5 \sqrt{3} m$

,which is the height of the tower.

• hope it helps you...
• please mark it as a brainlist answer...
• also please rate thanks and follow me....
• stay home STAY SAFE...