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 30-30-120, 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 30-60-90 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 30-60-90 triangle.
Given XO = radius =1
We want to the height of the triangle XPO which XP & opposite to angle 60 degree.
Applying 30-60-90 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.