Given a possible triangle ABC with a=5,b=11,and a=23, find a possible value for

The question is as follow:

Given a possible triangle ABC with a=5, b=11, and A=23 degrees, find a possible value for angle B. Round to the nearest tenth.
A. 10.2° B. 11.2° C. 59.3° D. no solution
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a = 5   ,    b = 11   ,   and    ∠A = 23°

By applying the Law of Sines:
[tex] \frac{sin \ A}{a} = \frac{sin \ B}{b} [/tex]

[tex] \frac{sin \ 23}{5} = \frac{sin \ B}{11} [/tex]

∴ sin B = 11* (sin 23°)/5 ≈ 0.8596

∴ ∠B = [tex] sin^{-1} 0.8596 [/tex] ≈ 59.3°

∴ The correct choice is option (C)
    C. 59.3°