Question

A rope dancer was walking on a loose rope tied to the top of two equal posts of

height 9 m. When he was at a certain height above the ground, it was found that the
stretched pieces of the rope made angles 30° and 60° with the horizontal line parallel
to the ground. If the total length of the rope is 20 m, how high was the position of
the rope dancer from the ground?​

Answer 1

Rope dancer is 2.66 meters above the ground.

Explanation:

In the figure attached,

Rope AB of length 20 meters has been tied between the poles A and B of height 9 meters.

When rope dancer is at point C, rope made 30° and 60° angles with the horizontal line AB.

Now we have to find the vertical distance of the rope dancer from the ground.

From ΔACD,

sin 30° =
`(h)/(x)`

h =
`(x)/(2)`

x = 2h

Similarly from ΔBCD,

sin 60° =
`(h)/((20-x))`

`(√(3))/(2)=(h)/((20-x))`

`(20-x)=(2h)/(√(3))`

x =
`20-(2h)/(√(3))`

Now by equating the values of x,

`2h=20-(2h)/(√(3) )`

`2h+(2h)/(√(3))=20`

`2h(1+(1)/(√(3)))=20`

`h=(10√(3))/((√(3)+1))`

h = 6.34 m

Now (9 - h) = 9 - 6.34

= 2.66 m

Therefore, rope dancer is 2.66 meters above the ground.