Find the angle B if in triangle ABC the angle is A = 30 degrees, AC = 2 cm, BC = √2.

It is known from the sine theorem that the sides of a triangle are related as the sines of their opposite angles. For this triangle, the equality is true:

AC / sin B = BC / sin A.

Hence, sin B = AC * sin A / BC = 2 * sin 30 / √2 = 2 * 0.5 / √2 = 1 / √2.

Find the angle B: B = arcsin (1 / √2) = 45 °.



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