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In triangle ABC, a=12.2, B=14.0, <A=43°, how do you find <B?

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Jun 2, 2015

Given a triangle ABC with
$XXXXX$ $color(white)("XXXXX")$ $∠ A = 43^{\circ}$ $/_A = 43^@$
$XXXXX$ $color(white)("XXXXX")$ $a = 12.2$ $a=12.2$
$XXXXX$ $color(white)("XXXXX")$ $b = 14.0$ $b=14.0$

We can use the Law of Sines
$XXXXX$ $color(white)("XXXXX")$ $\frac{sin (A)}{a} = \frac{sin (B)}{b}$ $(sin(A))/a = (sin(B))/b$

to determine a solution for $∠ B$ $/_B$
$XXXXX$ $color(white)("XXXXX")$ $∠ B = arcsin (\frac{b \cdot sin (A)}{a}) = {51.5}^{\circ}$ $/_B = arcsin((b*sin(A))/a) = 51.5^@$

BUT there are two possible solutions
$XXXXX$ $color(white)("XXXXX")$ as indicated in the diagram below:
enter image source here

The angle given by the Law of Sines is actually the angle for $∠ B_{2}$ $/_B_2$ in the diagram

$∠ B_{1} = 180^{\circ} - {51.5}^{\circ} = {128.5}^{\circ}$ $/_B_1 = 180^@ - 51.5^@ = 128.5^@$

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