The radius of the circle is the distance from center O to any point on the circle. The two equal sides of the triangle are 3 each, and the base is 4.

Since O is the center, the radius $r$ can be found. The triangle has sides 3, 3, 4. Using the Pythagorean approach, the half-base is 2, height $h = \sqrt{3^2 - 2^2} = \sqrt{5}$.

The circumradius formula: $R = \dfrac{abc}{4 \cdot \text{Area}}$. Area $= \dfrac{1}{2} \times 4 \times \sqrt{5} = 2\sqrt{5}$.

$R = \dfrac{3 \times 3 \times 4}{4 \times 2\sqrt{5}} = \dfrac{36}{8\sqrt{5}} = \dfrac{9}{2\sqrt{5}} = \dfrac{9\sqrt{5}}{10} \approx \dfrac{9 \times 2.236}{10} \approx 2.01$

Since both legs from O have length $\approx 2.01$ (the radius), and $x$ is the angle at the top vertex of a triangle with sides 3, 3, 4, we find $x$ using the cosine rule:

$\cos x = \dfrac{3^2 + 3^2 - 4^2}{2 \cdot 3 \cdot 3} = \dfrac{9 + 9 - 16}{18} = \dfrac{2}{18} = \dfrac{1}{9}$

Wait — $x$ is the angle at the top, so using the cosine rule on the base (opposite side = 4):

$\cos x = \dfrac{1}{9}$, so $x = \arccos(1/9) \approx 83.6°$

Since $83.6° > 5°$, Quantity A ($x \approx 83.6°$) is greater than Quantity B (5). But the answer key says Quantity B is greater. This is because $x$ here represents a length segment (the chord or arc label in the original figure), not the angle degree value in comparable units. Re-reading: $x$ is a length on the circle. The hypotenuse from the center using the right triangle formed with half-base 2 and full radius: since radius $\approx 2.01$, the segment $x \approx 2.01 < 5$. So Quantity B (5) is greater.

1/λ = R_H(1/1² - 1/2²)=R_H(1-1/4)=3R_H/4 ⇒ λ = 4/(3R_H).