Prime Angles

The sum of angle of any triangle is $180^\circ$ , so $\angle A+ \angle B + \angle C = 180^\circ$ .

Since we are given that $\angle A = 41^\circ$ , substituting it into the equation above and simplifying yields:

$\begin{aligned} \angle A+ \angle B + \angle C &=& 180^\circ \\ 41^\circ + \angle B + \angle C &=& 180^\circ \\ \angle B + \angle C &=& 180^\circ - 41^\circ \\ \angle B + \angle C &=& 139^\circ \\ \end{aligned}$

Since the $B$ and $C$ are are prime numbers and we know from above that the sum of these two prime numbers yields 139, which is an odd number, then by Parity properties of even and odd numbers , we have

$\text{ even } + \text{ odd } = \text{ odd }.$

In other words, one of the values of $B$ and $C$ has to be an even number. And because the only even prime number is 2, then either one of $B$ or $C$ is equal to 2 and the other is equal to $139 - 2 = 137$ . Since we are told that the angle $B$ is larger than the angle $C$ , then $B= 137$ and $C=2$ .

We can verify that all the numbers, $2,41,137$ are prime numbers using Sieve of Eratosthenes . Hence, all the conditions have been fulfilled. And so $B = \boxed{137}$ .