Question

I need help on this problem. I looked at an example but it it really confusing.
Solve the triangle....
a=1,b=4,C=60°
c=
A=
B=

Answer 1

cos theorem` c^2=a^2+b^2-2*a*b*cos(C)
so c^2=1+16-8*1/2
c^2=13
c=sqrt(13)

Answer 2

then, sin thorem a/sin(A)=b/sin(B)=c/sin(C)

Answer 3

so that is to find C,
and square root of 13 is 169

Answer 4

nor square of 13 is 169, square root of 13 is eqaul to about 3.6, but you can write it sqrt(13)

Answer 5

thats right, 3.6,
so next part is what I get stuck at,
cos^-1(1+13-4/4*1*square root of 13)

Answer 6

no you just need to solve it according to sin theorem
it is a/sin(A)=b/sin(B)=c/sin(C)
1/sin(A)=sqrt(13)/sin(60)
then sin(A)=sin(60)/sqrt(13)
and sin(B)=4*sin(60)/sqrt(13)

Answer 7

ok, is that to find A?

Answer 8

I am confused how to figure that part out?

Answer 9

if you found sin(A), that means you also found A, because then arcsin(sin(A))

Answer 10

ok, so to find A do I plug in what you wrote earlier?

Answer 11

yes

Answer 12

sin(60)=sqrt(3)/2
so sin(A)=sqrt(3)/(2*sqrt(13)
A=arcsin(sqrt(3)/(2*sqrt(13))