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Mathematics 13 Online
OpenStudy (anonymous):

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. B = 49°, a = 16, b = 14

OpenStudy (anonymous):

lol you all over the map with these huh?

OpenStudy (anonymous):

haha yeah

OpenStudy (anonymous):

Practice exam probably :P Do you know the law of sines?

OpenStudy (anonymous):

a/sin A = b/sin B = c/sin C

OpenStudy (anonymous):

Yeah, it's a practice exam

OpenStudy (anonymous):

\[\frac{\sin(49)}{14}=\frac{\sin(A)}{16}\]\[\sin(A)=\frac{16\sin(49)}{14}\]\[A=\arcsin(\frac{16\sin(49)}{14})\]

OpenStudy (anonymous):

i get about \(59.6\) http://www.wolframalpha.com/input/?i=arcsin%2816sin%2849%29%2F14%29

OpenStudy (anonymous):

now you can find angle C by making sure that the angles add to 180

OpenStudy (anonymous):

C = 71.4° ?

OpenStudy (anonymous):

that is what i get, yes

OpenStudy (anonymous):

You might only need to go so far due to some of the answers. But yes C=71.4

OpenStudy (anonymous):

Okay so C = 71.4° but what is c?

OpenStudy (anonymous):

yeah i guess now is the time to look at the choices since \(C=71.4\) you know \[\frac{c}{\sin(71.4)}=\frac{14}{\sin(49)}\]

OpenStudy (anonymous):

that makes \[c=\frac{14\sin(71.4)}{\sin(49)}\]

OpenStudy (anonymous):

our options: A = 59.6°, C = 71.4°, c = 11.1; A = 120.4°, C = 10.6°, c = 11.1 A = 59.6°, C = 71.4°, c = 17.6; A = 120.4°, C = 10.6°, c = 3.4

OpenStudy (anonymous):

So we don't need to solve for the other values once we find c we are done

OpenStudy (anonymous):

once we get c we don't have to do any more work

OpenStudy (anonymous):

ah I see :) thanks both of you

OpenStudy (anonymous):

You're welcome, and thank you even though I didn't do much

OpenStudy (anonymous):

yw you did fine

OpenStudy (anonymous):

Satellite should give his to her

OpenStudy (anonymous):

now you can help @girlfromafrica some more because i am old and tired and going to bed have fun!

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