# Help me with this ques.

• Feb 28th 2010, 02:30 PM
mrbains21
Help me with this ques.
Hey i have a question i cant solve it but m very curious to know its solution
Help me!!
Here it is:

Two rays are drawn through a point at an angle of 30*.A point B is taken on one of them at a distance d from the point A.A perpendicular is drawn from the point B to other ray and another perpendicular is drawn from as foot to meet AB at another point from where similar process is repeated indefinitely.
Calculate the length of the resulting infinite polygonal line.
• Feb 28th 2010, 04:05 PM
tonio
Quote:

Originally Posted by mrbains21
Hey i have a question i cant solve it but m very curious to know its solution
Help me!!
Here it is:

Two rays are drawn through a point at an angle of 30*.A point B is taken on one of them at a distance d from the point A.A perpendicular is drawn from the point B to other ray and another perpendicular is drawn from as foot to meet AB at another point from where similar process is repeated indefinitely.
Calculate the length of the resulting infinite polygonal line.

I suppose point A is the intersection point of both rays...then, after choosing point B on ray 1 and drawing from it a perpendicular to ray 2 you get a straight-angle triangle 30-60-90, sometimes aka golden triangle, with hipotenuse d, and thus the leg opposite to the 90 deg. angle is half the hipotenuse ==> the first part of the polygonal line is $\frac{1}{2}d$, and the other leg's length is, by Pythagoras, $\frac{\sqrt{3}}{2}d$. Repeat the process, again you get a 30-60-90 triangle but with hipotenuse $=\frac{\sqrt{3}}{2}d$, so this time the polygonal line's length is $\frac{\sqrt{3}}{4}d$ , and etc.

The whole polygonal line's length is thus $\sum\limits_{k=0}^\infty \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)^k=\frac {\frac{1}{2}}{1-\frac{\sqrt{3}}{2}}$ $=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$

Tonio
• Mar 1st 2010, 03:46 AM
mrbains21
Quote:

Originally Posted by tonio
I suppose point A is the intersection point of both rays...then, after choosing point B on ray 1 and drawing from it a perpendicular to ray 2 you get a straight-angle triangle 30-60-90, sometimes aka golden triangle, with hipotenuse d, and thus the leg opposite to the 90 deg. angle is half the hipotenuse ==> the first part of the polygonal line is $\frac{1}{2}d$, and the other leg's length is, by Pythagoras, $\frac{\sqrt{3}}{2}d$. Repeat the process, again you get a 30-60-90 triangle but with hipotenuse $=\frac{\sqrt{3}}{2}d$, so this time the polygonal line's length is $\frac{\sqrt{3}}{4}d$ , and etc.

The whole polygonal line's length is thus $\sum\limits_{k=0}^\infty \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)^k=\frac {\frac{1}{2}}{1-\frac{\sqrt{3}}{2}}$ $=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$

Tonio

when i solved it i got d*(2+sqrt(3))
• Mar 1st 2010, 04:50 AM
tonio
Quote:

Originally Posted by mrbains21
when i solved it i got d*(2+sqrt(3))

Of course: I just forgot to multiply by d all along, but d appears in my post , too.

Tonio