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## A circle has center at origin and radius 1......

tagged by: Vincen

This topic has 1 expert reply and 0 member replies

### Top Member

Vincen Legendary Member
Joined
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#### A circle has center at origin and radius 1......

Wed Jan 24, 2018 8:33 am
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)Ï€.What is the length of line segment XZ?

A. âˆš2
B. âˆš3
C. 2
D. 2+âˆš3
E. 2+âˆš2

The OA is the option B.

How can I solve this PS question? I don't know how to find the length of XZ? Experts, could you help me? Thanks.

### GMAT/MBA Expert

EconomistGMATTutor GMAT Instructor
Joined
04 Oct 2017
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551 messages
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10 members
180
Fri Jan 26, 2018 11:35 am
Quote:
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)Ï€.What is the length of line segment XZ?

A. âˆš2
B. âˆš3
C. 2
D. 2+âˆš3
E. 2+âˆš2

The OA is the option B.

How can I solve this PS question? I don't know how to find the length of XZ? Experts, could you help me? Thanks.
Hi Vincen,
Let's take a look at your question.

Center of the circle is (0, 0).

We will use the angle XOZ first using the arc length formula.
$$S=\ r\left(theta\right)$$ $$\frac{2\pi}{3}=1\left(theta\right)$$
$$theta=\frac{2\pi}{3}$$

Now we can find the length of XZ using cosine formula in triangle XOZ.
The cosine formula is:
$$a^2=b^2+c^2-2bc\ Cos\left(A\right)$$
Where A is the angle opposite to the side length a and between the sides b and c.
$$\left(XZ\right)^2=\left(1\right)^2+\left(1\right)^2-2\left(1\right)\left(1\right)Cos\left(\frac{2\pi}{3}\right)$$
$$\left(XZ\right)^2=1+1-2\left(-\frac{1}{2}\right)$$
$$\left(XZ\right)^2=2+1$$
$$\left(XZ\right)^2=3$$
$$XZ=\sqrt{3}$$

Therefore, option B is correct.

Hope it helps.
I am available if you'd like any follow up.

_________________
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