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A circle has center at origin and radius 1. The points X, Y
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25 Aug 2012, 21:04
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A circle has center at origin and radius 1. The points X, Y, and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ? A. √2 B. √3 C. 2 D. 2+√3 E. 2+√2 I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method?
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Re: A circle has center at origin and radius 1.The points X,Y,
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26 Aug 2012, 00:43
vinay911 wrote: A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?
a)√2
b)√3
c)2
d)2+√3
e)2+√2
I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method? If the arc XYZ is \(2/3\pi\), the central angle XOZ is \(120^o\) (2/3*360=120). Then you can calculate XZ from the isosceles triangle XOZ which has angles 3030120, and XO = OZ = 1 = radius. The height to the base of the triangle is 0.5 (opposing the \(30^o\) angle), so half of the base is \(0.5*\sqrt{3}{\), and \(XZ=\sqrt{3}.\) Answer B.
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Re: A circle has center at origin and radius 1. The points X, Y
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26 Aug 2012, 08:56
Thanks Eva. I have a clarification here: I am not sure how you got the length of the perpendicular to side XZ as 0.5.Is it by trigonometry(sin 30 ?) ? Also how do you say the perpendicular from the center divides XZ equally ?



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Re: A circle has center at origin and radius 1. The points X, Y
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26 Aug 2012, 08:59
vinay911 wrote: Thanks Eva. I have a clarification here: I am not sure how you got the length of the perpendicular to side XZ as 0.5.Is it by trigonometry(sin 30 ?) ? Also how do you say the perpendicular from the center divides XZ equally ? Yes, is "hidden" trigonometry... In a right triangle of type 306090 the leg opposed to the 30 degrees angle is half of the hypotenuse. Good to remember: in a right triangle of type \(306090\) the sides are \(xx\sqrt{3}2x\).
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Re: A circle has center at origin and radius 1. The points X, Y
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26 Aug 2012, 09:14
Gotcha! i always knew this but din't strike me here +1 EvaJ and you hit the half century kudos mark



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Re: A circle has center at origin and radius 1.The points X,Y,
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07 Sep 2012, 12:55
EvaJager wrote: vinay911 wrote: A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?
a)√2
b)√3
c)2
d)2+√3
e)2+√2
I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method? If the arc XYZ is \(2/3\pi\), the central angle XOZ is \(120^o\) (2/3*360=120). Then you can calculate XZ from the isosceles triangle XOZ which has angles 3030120, and XO = OZ = 1 = radius. The height to the base of the triangle is 0.5 (opposing the \(30^o\) angle), so half of the base is \(0.5*\sqrt{3}{\), and \(XZ=\sqrt{3}.\) Yes, is "hidden" trigonometry... In a right triangle of type 306090 the leg opposed to the 30 degrees angle is half of the hypotenuse. Answer B. Hi Eva, I would like to add few intermediate steps to the solution offered by you. As Eva has correctly pointed that the central angle XOZ is \(120^o\) Now drop a perpendicular from point X & extend the line OZ in the direction of that perpendicular & name that meeting point P If you draw such triangle then you will that triangle XPO is 306090 triangle. Given XO = radius =1 We want to the height of the triangle XPO which XP & opposite to angle 60 degree. Applying 306090 theorem , we get XP = \sqrt{3}/2 and this height is the height of the original triangle XOZ Property 1  The Altitude to the base of an isosceles or equilateral triangle bisects the base & bisects the vertex angle.Property 2  The midpoint of the hypotenuse is equidistant from the three polygon vertices Because the height (altitude) bisects the side XZ, thus is equidistant from X,O or Z Therefore XZ = 2 x height = 2 \sqrt{3}/2 = \sqrt{3} Answer B I hope this extra long explanation will help many & sorry that i didn't include the possible diagram.
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Re: A circle has center at origin and radius 1.The points X,Y,
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24 Aug 2013, 22:05
EvaJager wrote: vinay911 wrote: A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?
a)√2
b)√3
c)2
d)2+√3
e)2+√2
I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method? If the arc XYZ is \(2/3\pi\), the central angle XOZ is \(120^o\) (2/3*360=120). Then you can calculate XZ from the isosceles triangle XOZ which has angles 3030120, and XO = OZ = 1 = radius. The height to the base of the triangle is 0.5 (opposing the \(30^o\) angle), so half of the base is \(0.5*\sqrt{3}{\), and \(XZ=\sqrt{3}.\) Answer B. Sorry EvaJager, in fact, I agree with the result that the central angle XOZ is 120 degree. However, I am afraid that the equation 2/3 * 360 = 120 is not correct (in fact 240). The result of 120 degree of central angle XOZ can be explained as follows: The arc XYZ is 2/3pi . The circumference of circle (O,1) is 2pi. Therefore, arc XYZ makes up 1/3 of circle (O,1). And the result that the central angle XOZ is 1/3 * 360 = 120 degree can be inferred.



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Re: A circle has center at origin and radius 1.The points X,Y,
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24 Aug 2013, 22:57
EvaJager wrote: vinay911 wrote: A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?
a)√2
b)√3
c)2
d)2+√3
e)2+√2
I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method? If the arc XYZ is \(2/3\pi\), the central angle XOZ is \(120^o\) ( 2/3*360=120). Then you can calculate XZ from the isosceles triangle XOZ which has angles 3030120, and XO = OZ = 1 = radius. The height to the base of the triangle is 0.5 (opposing the \(30^o\) angle), so half of the base is \(0.5*\sqrt{3}{\), and \(XZ=\sqrt{3}.\) Answer B. Above  2/3*360=120 should be 2/3*180=120.
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Re: A circle has center at origin and radius 1.The points X,Y,
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25 Aug 2013, 01:20
EvaJager wrote: vinay911 wrote: A circle has center at origin and radius 1.The points X,Y,and Z lie on the circle such that the length of arc XYZ is (2/3)π.What is the length of line segment XZ?
a)√2
b)√3
c)2
d)2+√3
e)2+√2
I figured we could eliminate c,d,e based on the rule that for a triangle,the 3rd side must be greater than difference of other 2 sides & lesser than the sum of other two sides.I am stuck after that. is there an neater method? If the arc XYZ is \(2/3\pi\), the central angle XOZ is \(120^o\) (2/3*360=120). Then you can calculate XZ from the isosceles triangle XOZ which has angles 3030120, and XO = OZ = 1 = radius. The height to the base of the triangle is 0.5 (opposing the \(30^o\) angle), so half of the base is \(0.5*\sqrt{3}{\), and \(XZ=\sqrt{3}.\) Answer B. i know this is incorrect but please clarify, if the triangle was isoceles , then XZ is [square_root]2 *1 = [square_root]2



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Re: A circle has center at origin and radius 1. The points X, Y
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11 Jul 2016, 05:20
Use the formula:
Radius of circumscribed circle = side * (sq. root 3 / 3 )
We have an equilateral triangle inscribed in a circle. (confirmed by the fact that the angle corresponding to 2pi/3 is 120º) So we can use the above formula.



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Re: A circle has center at origin and radius 1. The points X, Y
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