Funny trig paradox...I think

[Dionysus22]

Small question:

A triangle abc has sides |ab| = 7, |bc| = 3, and |ac| = 8. If the angle |<acb| is 60 (degrees),

Find the other two angles in the triangle.

Now, I've used the Sine Rule, but I'm getting two different answers.

If I use (|ab|/Sin60) = (|bc|/SinX) I get different answers to what I get if I use (|ab|/Sin60) = (|bc|/SinX).

Any pointers??

[mmcc79]

I don't know enough to say why that will not work, but I would advise you to use the following formula when you know the three sides of a triangle...

c^2 = a^2 + b^2 - 2*a*b*Cos C

Where C would be the angle <acb.

Then switch c with a... and use Cos A (<cab)
Then switch c with b... and use Cos B (<cba)

[HallsofIvy]

"If I use (|ab|/Sin60) = (|bc|/SinX) I get different answers to what I get if I use (|ab|/Sin60) = (|bc|/SinX)."

?? Those are exactly the same! What do you expect to get?

If you mean you get two angles whose some is not 120 degrees (so that the sum of the three angles is not 180), remember that sin(A)= x has TWO solutions between 0 and 180 degrees. The "principle" root is between 0 and 90 degrees. To find the other, subtract that answer from 180 degrees.