How do you use the law of sines to solve the triangle given A = 165.1°, B = 81.9°, b = 18.6?

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How do you use the law of sines to solve the triangle given A = 165.1°, B = 81.9°, b = 18.6?

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Answer 1 · 2015-05-28T20:39:01

The law of sines says that in a triangle where the angle $α$ $alpha$ is opposite the side of length $a$ $a$ , angle $β$ $beta$ - length $b$ $b$ and angle $γ$ $gamma$ - length $c$ $c$ and $R$ $R$ is the radius of the circumscribed circle (the smallest circle possible that contains the triangle / passes through all three vertices) we have the following:
$\frac{a}{sin α} = \frac{b}{sin β} = \frac{c}{sin γ} = 2 R$ $a/sin alpha = b/sin beta = c/sin gamma = 2R$
and you can use any two of these, e.g. $\frac{b}{sin β} = 2 R$ $b/sin beta=2R$ .

In your question values $A$ $A$ and $B$ $B$ are given in degrees so one can assume that they're the angles $α$ $alpha$ and $β$ $beta$ but their sum is $247^{\circ}$ $247^@$ which is a little bit too much since in all triangles the sum of all three angles is always $180^{\circ}$ $180^@$ . Please check where the mistake in those values is :)

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How do you use the law of sines to solve the triangle given A = 165.1°, B = 81.9°, b = 18.6?

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