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

1 Answer

Jun 2, 2015

Given a triangle ABC with
$color(white)("XXXXX")$ $/_A = 43^@$
$color(white)("XXXXX")$ $a=12.2$
$color(white)("XXXXX")$ $b=14.0$

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

to determine a solution for $/_B$
$color(white)("XXXXX")$ $/_B = arcsin((b*sin(A))/a) = 51.5^@$

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

The angle given by the Law of Sines is actually the angle for $/_B_2$ in the diagram

$/_B_1 = 180^@ - 51.5^@ = 128.5^@$

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