Question

Solve the triangle
A=50 degrees, b=13, c=6

a=10.2,C=26.3,B=103.7
a=14,C=26.3,B=103.7
No triangles possible
a=14,C=30.3,B=99.7

Answer 1

law of sines

Answer 2

I think you'll need law of cosines with that information.

Answer 3

the question asks for law of sines

Answer 4

Law of sines looks like this
\[\large \frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}\]
Right now, all you have is
\[\large \frac{a}{sin(50º)}=\frac{13}{sin(B)}=\frac{6}{sin(C)}\]

Answer 5

It's possible with a little algebraic juggling, but I still think it's easier to use law of cosines first to get the third side, then law of sines will clean the rest up nicely.

Answer 6

The law of cosines is:

\(c^2 = a^2 + b^2 + 2ab \cdot cos C\)

Other equations:

\(a^2 = b^2 + c^2 - 2bc \cdot cos A\) <--- This one is the one you probably want.
\(b^2 = a^2 + c^2 - 2ac \cdot cos B\)

http://en.wikipedia.org/wiki/Law_of_cosines

Answer 7

If you try it with law of sines alone, you'll need to set up a system of three equations for the three unknowns, and that will get a little annoying.

Answer 8

i have to choose from the options above

Answer 9

I still recommend finding a first, then you can use law of sines to get B and C.

Answer 10

okay thankyou

Answer 11

You're welcome.