0) In \( \triangle X Y Z, \frac{2 x}{\sin x}=\ldots \quad \mathrm{r} \) "where r is the radius length of the ircumference of \( \triangle X Y Z^{\prime \prime} \) \( \begin{array}{llll}\text { a) } 1 & \text { b) } 2 & \text { c) } 4\end{array} \)

In triangle \( XYZ \), the relationship given is \[ \frac{2x}{\sin x} = r \] where \( r \) is the radius of the circumcircle of triangle \( XYZ \). To find the correct answer, we can use the formula for the circumradius \( R \) of a triangle, which is given by: \[ R = \frac{abc}{4K} \] where \( a, b, c \) are the lengths of the sides of the triangle, and \( K \) is the area of the triangle. However, the relationship you provided seems to be a specific case. In general, for any triangle, the circumradius \( R \) can also be expressed in terms of one of its angles and the opposite side: \[ R = \frac{a}{2\sin A} \] where \( a \) is the length of the side opposite angle \( A \). If we assume \( x \) corresponds to angle \( A \) and \( 2x \) corresponds to the side opposite angle \( A \), we can rewrite the relationship as: \[ R = \frac{2x}{\sin x} \] This suggests that \( r \) is indeed equal to \( R \). Now, we need to determine the value of \( r \) based on the options provided: - a) 1 - b) 2 - c) 4 Since the circumradius \( R \) can take various values depending on the specific triangle, we cannot definitively choose one of the options without additional context or constraints on the triangle. However, if we consider a standard triangle where the circumradius is often normalized or simplified, the most common value for \( R \) in many contexts is \( 2 \) (especially in the case of a right triangle or an equilateral triangle). Thus, the most reasonable answer based on common geometric principles would be: \[ \text{b) } 2 \]