Decide whether the Law of Sines or the Law of Cosines can be used to find the measure of c
For law of sines- you need to know one side and other angles. law of cosines- two sides and including angle. Can you try now?
praly but idk how
Do you know the laws?
ehh kinda sorta
Okay, Law of sines-\[\frac{ a }{ \sin A }=\frac{ b }{ \sin B }=\frac{ c }{ sinC } =2R\] Law of cosines-\[a^2 = b^2+c^2 - 2bc \cos A\] where , a,b,c- lengths of sides of triangles A,B,C- angles R- circumradius of triangle. Clear?
nope
Then first learn the laws properly from somewhere.
can u show me how to do my problem though
If you can understand the equations, In the problem we know the angles and ONLY ONE side. So, cosine rule cannot be applied, since we need another side in it. We can use sine rule here.
can u set it up
Okay, From sine rule \[\frac{ c }{ \sin C} = \frac{ b }{ sinB }\] \[\frac{ c }{ \sin 40 }=\frac{ 18 }{ \sin 115 }\] Now we can get 'c'.
ok
Fine, try understanding the laws properly before any further problems..
I CANT DO ANY OF THE PROBLEMS ANYWAY!!!!!!!!!!!!!!!! if i understood n didnt need help i wouldnt ASK
@koushik_ksv there is absolutely no reason to be a totes jerk about it he needs he;p understanding something he doesn't understand, why are you being mean about it? don't answer if your gonna be rude and impatient
he isnt bein a jerk ry i just dont understand anything he puts up there
.-. then tf
Folks, Koushik has just taken some time to work through the problem. IF it is still not clear, the right thing to do is to ask another question so he can help or ask for someone else to help. Calling him names is completely unjustified.
I tried my best to make him understand the laws. I would be definitely happy if some one else can do it better. It would benefit him, not me. @Ryleighblue
i need to understand how to do my question not the laws
When you are perfect with the laws, you can solve many more questions easily. This question would be very easy for you. Anyway, I have shown you how to solve this. If you have any doubts, you can ask me.
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