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?

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eudora eudora · Answer 1 · 2019-12-14T19:07:02+01:00

Answer:

Rope dancer is 2.66 meters above the ground.

Step-by-step 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° = [tex]\frac{h}{x}[/tex]

h = [tex]\frac{x}{2}[/tex]

x = 2h

Similarly from ΔBCD,

sin 60° = [tex]\frac{h}{(20-x)}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{h}{(20-x)}[/tex]

[tex](20-x)=\frac{2h}{\sqrt{3}}[/tex]

x = [tex]20-\frac{2h}{\sqrt{3}}[/tex]

Now by equating the values of x,

[tex]2h=20-\frac{2h}{\sqrt{3} }[/tex]

[tex]2h+\frac{2h}{\sqrt{3}}=20[/tex]

[tex]2h(1+\frac{1}{\sqrt{3}})=20[/tex]

[tex]h=\frac{10\sqrt{3}}{(\sqrt{3}+1)}[/tex]

h = 6.34 m

Now (9 - h) = 9 - 6.34

= 2.66 m

Therefore, rope dancer is 2.66 meters above the ground.