Billy J.

^{-1}(1.0244) which is a domain error does that mean the answer is it is not a valid triangle?

04/02/16

Billy J.

asked • 04/02/16I have

a = 65

b = ?

c = 50

<A = ?

<B = ?

<C = 52

Now I understand the b^{2} = a^{2} + c^{2} - 2ac*CosB and the other 2 variants of the formula but how do I solve it when I don't have a value to put for Cos B in the formula above?

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Billy J.

So I did that and I ended up with sin^{-1}(1.0244) which is a domain error does that mean the answer is it is not a valid triangle?

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04/02/16

You are given the lengths of side a and c. To use the Law of Cosines you wold need to know m∠B.

Try the Law of Sines.

Billy J.

So I did that and I ended up with sin-1(1.0244) which is a domain error does that mean the answer is it is not a valid triangle?

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04/02/16

Billy J.

sin^{-1}(1.0244)** is what I meant

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04/02/16

Mark M.

tutor

sin 52° / 50 = sin A / 65

0.7880 / 50 ≈ sin A / 65

51.22 / 50 ≈ sin A

1.02 ≈ sin A

Either not a valid triangle, or the information is inaccurate.

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04/02/16

Billy J.

Thank you for the help

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04/02/16

Mark M.

tutor

You are welcome!

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04/02/16

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Billy J.

04/02/16