can anybody solve this?

Question

A circular park of radius 20m is situated in a colony.Three boys Ankur,Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other.Find the length of the string of each phone?
Please email me the figure at
My email id is [e-mail]

Answer 1

See the diagram here. - http://img246.imageshack.us/my.php?image=problemem4.gif

In the Diagram, A,B and C are the boys. AB, BC, AC are the strings of phones.

O is the centre and COP is a line passing through centre and making a 90 degree angle with AB. Now AP = BP = x.

And the line BO, bisects the angle ABC into two equal halves. So angle PBO is 30 degrees.

Now consider the right angled triangle PBO.

In this triangle, cos B = BP / BO

that is cos 30 = x / 20

that is x = 20 cos 30 = 17.32 m

Since AP = BP = x,

AB = 2x

AB = 34.64 m

So the length of each string is 34.64 m

Answer 2

Draw your own diagram.

A, S and D are at the vertices of an inscribed equilateral triangle with centroid the circle's centre, C.

Consider the triangle ACD, with enclosed angle 120 deg. (by symmetry). Drop a perpendicular from C onto AS. You now have two similar right-angled triangles, and need only consider one of them, say the one with hypoteneuse AC. Length AC = 20m. Therefore the length of the little perpendicular you just dropped is 10m (the half-angle at C is 60 deg.).*** Pythagoras now implies that HALF of AS is 10 (sqrt 3) m, so the distance AS, presumably what you're after, is 20 (sqrt 3) m.

Live long and prosper.

*** You could use the cosine formula with angle ACD = 120 deg. to find AS directly. However, with an angle like 60 deg. identifiable in a problem, I always find it easiest to remember that the projection from a 60 deg. sloping line onto the relevant "base" is just half the sloping line's length. I much prefer using that to bothering with "now let's see, is this sine or cos, and what's that value anyway," or equivalently remembering that the other perpendicular factor is (sqrt 3)/2 or "let's see again, which of sine or cos do I need this time and etc. ...?"). But then, we all have our own peculiar little ways of calling up useful results.

Answer 3

Each cord intersect an angle of 360/3=120°
this givs a triangle with angles of 120°, 30° & 30°
20m/sin 30=x/sin 120
40=x/sin 120
x=40=sin 120=34.64meters.

Answer 4

Make a scale of 1 mm for 1 m, 20 mm for 20 Mts.

NOw draw the circle of 20 mm radius. draw the tangents to form the triangle outside the Circle. The point of intersection of Line and Circle are marked. Join all the points. Total distanceof this point is length and multiply by 1000 to get the answer.

Answer 5

You have an isosceles triange with central angle of 120 degrees, and adjacent sides of 20 meters. The length of the third side is 20 times the square root of three.

Answer 6

if the length is A = B = C
A^2 = 20^2 + 20^2 - 2*20*20*cos120
A^2 = 40 + 40 - 800*(-1/2)
A^2 = 80 + 400
A^2 = 480
A = 21.91 m
: )

Answer 7

its a equilateral triangle inside a circle of 20 m radius.whose one side is equal to 34.62m

Answer 8

(20cos30) x 2

Answer 9

20sqrt3

Answer 10

may b 60m no idea this answer may be wrong